A 20.0 N block slides over a horizontal table. If it takes 5.0 N to move the block at a constant velocity, the coefficient of friction is D. 0.25.
Given,
Frictional force (F) is 5.0 N
Normal force (N) is 20.0 N
The coefficient of friction (µ), a numerical value, is obtained by dividing the resistive force of friction (F) by the normal or perpendicular force (N) pushing the objects together.
i.e., µ = F/N
Calculating the coefficient of friction :
The frictional force is equal to and opposing the applied force since the box is traveling at a constant speed.
µ = F / N
µ = 5.0 N / 20.0 N
µ = 1/4 N
µ = 0.25 N
Thus, the coefficient of friction is D. 0.25.
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If Mercury's average orbital distance from the sun is 0.312 AU and k for our sun is 1.00 AU3/yr2, then what is Mercury's orbital period?
Take into account that by the Kepler's Third Law, you can use the following formula:
[tex]k^{}=\frac{a^3}{T^2}[/tex]where,
T: period of Mercury
a: average distance from Mercuty to Sun = 0.312AU
k: Kepler's constant = 1.00 AU^3/yr^2
Then, by replacing the previous values of the parameters, you obtain for T:
[tex]T=\sqrt[]{\frac{a^3}{k}}=\sqrt[]{\frac{(0.312AU)^3}{1.00\frac{AU^3}{yr^2}}}\approx0.56yr[/tex]Hence, the Mercury's orbital period is approximately 0.56 yr
Place the red, 100 g ball at the 60 cm mark on the ramp. How far did the box move? a90 cm b0cm cno more that 100 cm
Answer:
place it on the 50 cm mark
The gravitational force between two masses is 36 N. What is the gravitational force if the distance between them is tripled? (G = 6.673 x10 (Power of-11)
If the distance between them is tripled, the gravitational force is F2= 36/9=4N. The force of attraction between any two bodies is proportional to the product of their masses and inversely proportional to the square of their distance.
WHAT IS GRAVITATIONAL FORCE?
It is the force that connects all masses in the universe, particularly the pull of the earth's mass on objects near its surface. It follows the inverse square law. The force of attraction exerted on a body by the earth is known as gravitational force. For example, the leaves and fruits of a tree fall to the ground, water in a river flows down streams, and a ball thrown up travels to a height before returning to the ground are all examples of motion caused by gravitational force.The formula for Gravitational force:
F=G m1*m2/r² ,
G: gravitational constant = 6.67x10⁻¹¹ N*m²/kg²
m1: is the mass of the first object
m2: is the mass of the second object
r: is the distance between the center of the masses of the objects
F∝1/r²
so F1/F2=( r2/r1)².......(1)
Given F1= 36N F2=?
let r1=r, r2= 3r
putting all this value in equation (1)
36/F2= (3r/r)²
⇒F2= 36/9=4N
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Find the answer to the problem
The box is moving at a 2.06 m/s rate.
What causes acceleration?Acceleration is the rate at which velocity changes over time in both the direction and the speed. Anything is said to have been accelerated when it goes faster or slower in a straight line. Even travel on a circle accelerates steadily since the direction is continuously changing.
What metrics are used to quantify acceleration?Acceleration is measured in meters per second per second (m/s2).
Formulas for
F NET are F NET = m and F NET = F1 + F2.
3 + 3 + 7 Equals 13 when F NET is used.
F NET = m * a then,
where a = F NET/m and
a = 13N / 6.3 kg and 2.06m/s respectively.
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The velocity-time table represents the motion of a leftward moving car.
What is the magnitude and direction of the acceleration?
Given velocity-time table represents the motion of a leftward moving car.
Magnitude of acceleration = -4 m/s^2
Direction is Left.
When any object is slowing down then the acceleration is in opposite direction. Then the object will have a negative acceleration.
Deceleration is the opposite of acceleration. The deceleration is calculated by dividing the difference final and initial velocity by the amount of time taken for the drop in velocity. The formula for acceleration with a negative sign is used to identify the deceleration value.
Deceleration = (Final Velocity–Initial Velocity)/Time taken
It is written' –a,' where a is acceleration.
When initial velocity, final velocity and time are given, then Deceleration Formula is given by,
Given in the graph, u= 20m/s and v= 4m/s
t= 4sec
a = (v−u)/t
= (4 -20)/ 4
=- 16/4
a= -4 m/s^2
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Solving For Temperature (T)
7. How many degrees would the temperature of a 450 g piece of iron increase if 7600 J of
energy are applied to it? (The specific heat of iron is0.4494 J/g x °C)
Temperature is a unit used to represent how hot or cold something is. It can be stated using the Celsius or Fahrenheit scales, among others. Temperature shows which way heat energy will naturally flow, i.e., from a hotter (body with a higher temperature) to a colder (body with a lower temperature) (one at a lower temperature)
Additionally, we are provided the iron's particular heat as well as the iron skillet's mask.
We must determine the required quantity of heat based on this information. It will be Q = c x mass x delta P using this method.
That is the temperature change. Therefore, we can replace the value here. The c represents the iron's particular temperature. That comes to 0.4499.
The supplied mass is 5.07, and the resulting temperature change will be 250 - 25 °C. If we can solve this, we can determine what Q is worth.
Alternately, we might state that the required amount of heat is 512.19 Joules.
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A car accelerates from rest at -1.60 m/s2. (Include the sign in your answers.)(a) What is the velocity at the end of 5.6 s? m/s(b) What is the displacement after 5.6 s?
Given data
*The given initial velocity of the car is u = 0 m/s
*The given acceleration of the car is
[tex]undefined[/tex]Energy stored in the kinetic energy container is fundamentally different from energy stored in the elastic energy container.True or false
The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]Where:
m = mass of the object
v = Speed of the object
The elastic potential energy is given by:
[tex]U=\frac{1}{2}k\Delta x[/tex]Where:
k = Constant of the spring
Δx = Displacement
As we can see, both energy depends on different factors, therefore we can conclude the statement is true
Answer:
True
L../IN10. An object of mass 60.986 kg is sliding on a horizontalsurface with a uniform speed. The coefficientof kinetic fiction of the surfaces is 0.15. Calculatethe force of friction exerted by the surface on the object. (1 point)A. O 101.798 NB. O114.077 NC. 050.2 ND. 145.451 NE. O 89.649 N
Given:
The mass of the object is,
[tex]m=60.986\text{ kg}[/tex]The coefficient of friction is,
[tex]\mu=0.15[/tex]The object is sliding on the horizontal surface at a uniform speed.
To find:
the force of friction exerted by the surface on the object
Explanation:
If we draw the free body diagram of the object we see,
The normal reaction is R, the applied force is F, and the frictional force isf.
As the object is continuing its constant speed, we can say the object is in equilibrium.
So,
[tex]R=mg[/tex]and
[tex]\begin{gathered} f=F=\mu R \\ f=\mu mg \\ f=0.15\times60.986\times9.8 \\ f=89.649\text{ N} \end{gathered}[/tex]Hence, the friction force exerted by the surface is 89.649 N.
a lot of this content is conceptual but I'm having a hard time understanding this
If momentum is denoted by p and wavelength by
[tex]\lambda[/tex]Then, the wavelength and momentum are related as
[tex]p\propto\frac{1}{\lambda}[/tex]Thus if momentum is doubled, then the wavelength is halved.
An automobile with an initial speed of 4.30 m/s accelerates uniformly at the rate of 3.0 m/s/s. Find the final speed and the displacement after 5.0s
The final speed of the automobile with an initial speed of 4.30 m/s that rates uniformly at the rate of 3.0 m/s/s for 5 s is 19.3 m / s and its displacement will be 59 m
v = u + at
v = Final velocity
u = Initial velocity
a = Acceleration
t = Time
u = 4.3 m / s
a = 3 m / s / s
t = 5 s
v = 4.3 + ( 3 * 5 )
v = 19.3 m / s
s = ut + 1 / 2 at²
s = Displacement
s = ( 4.3 * 5 ) + ( 1 / 2 *3 * 5 * 5 )
s = 21.5 + 37.5
s = 59 m
Therefore,
Final speed = 19.3 m / sDisplacement = 59 mTo know more about Displacement
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Identify the time interval during which the velocity of shuttle bus is zero.
Explanation:
during D and E, so between 13 and 16 (whatever time unit this is - hours, minutes, ... ?).
Answer:
Explanation:
The time interval 13 to 16 seconds
A soccer ball kicked with a force of 13.5 N accelerates at 6.5 m/s^2. What is the mass of the ball?
ANSWER:
2.08 kg
STEP-BY-STEP EXPLANATION:
The force is given by the multiplication of the mass and the acceleration, like this:
[tex]\begin{gathered} f=m\cdot a \\ f=13.5\text{ N} \\ a=6.5\frac{m}{s^2} \end{gathered}[/tex]We replace and calculate for m:
[tex]\begin{gathered} 13.5=m\cdot6.5 \\ m=\frac{13.5}{6.5} \\ m=2.08\text{ kg} \end{gathered}[/tex]The mass is 2.08 kg
9. How much change in temperature would the addition of 35 000 Joules of heat have ona
538.0 gram sample of copper? (Look up specific heat of copper)
The change in temperature would be 168.97 °C
Q = cm∆t
Where
Q = Amount of heat energy
c = specific heat capacity of copper metal = 0.385 J/g°C
m = mass of copper
∆t = change in temperature
From the question,
Given:
m = 538 g
c = 0.385 J/g°C
Q = 35 000 Joules
Q = cm∆t
35000 = 0.385 x 538 x ∆t
∆t = 168.97 °C
Hence, change in temperature would be 168.97 °C
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Two identical metallic spheres A and
B, each carrying a charge of q C, are fixed.
They repel each other with a force of
2x10-5 N. Another identical uncharged
sphere C is made to touch A, moved to
touch B, and placed halfway between A
and B. What is then the electric charge
and the electric force (in N) on each
sphere in terms of q?
The electric charge and the electric force on each sphere in terms of q is 2
The Electric Charge and Electric ForceThe ability of particles or objects to attract or repel one another without coming into contact is known as electric charge. Incompatiblely charged particles are drawn to one another. Like-charged particles repel one another. Electric force is the name given to the force of attraction or repulsion.
Two metallic spheres, a and b, with a charge of half and a force of two tens of five
F1=k(1/2)(1/2)/r2
Equation F1=k/4r
A and C are two metallic spheres, respectively, so
Equation F2=k/2rsquare
equ1/equ2
2×10^-5/F2=k/4r^2×2r^2/k
F2=4×10^-5
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Exactly 3.3 s after a projectile is fired into the air from the ground, it is observed to have a velocity v⃗ = (8.6 i^ + 4.7 j^)m/s, where the x axis is horizontal and the y axis is positive upward. aDetermine the horizontal range of the projectile.bDetermine its maximum height above the ground.cDetermine the speed of motion just before the projectile strikes the ground.dDetermine the angle of motion just before the projectile strikes the ground.= ∘ below the horizontal
a. Horizontal range is given by,
[tex]Range=\frac{u^2sin2\theta}{g}[/tex]where u is the initial speed, and theta is the angle of projection,
Now velocity vector as a function of time is given by,
[tex]\vec{v}(t)=(ucos\theta)\hat{i}+(usin\theta-gt)\hat{j}[/tex]Given,
[tex]\begin{gathered} \vec{v}(3.3s)=8.6\text{ }\hat{i}+4.7\text{ }\hat{j} \\ \Rightarrow ucos\theta=8.6,usin\theta-(10)(3.3)=4.7 \\ \Rightarrow ucos\theta=8.6,usin\theta=37.7 \\ Dividing, \\ \Rightarrow tan\theta=\frac{37.7}{8.6}=4.3837 \\ \Rightarrow\theta=77.15\degree \\ Hence \\ ucos77.15\degree=8.6 \\ \Rightarrow u=\frac{8.6}{0.2224}=38.67\text{ m/s} \end{gathered}[/tex]And so, the horizontal range will be,
[tex]Range=\frac{(38.67)^2sin(154.3\degree)}{(10)}=64.85\text{ m}[/tex]b. Maximum height reached by a projectile is given by,
[tex]\begin{gathered} H_{max}=\frac{u^2sin^2\theta}{2g} \\ \Rightarrow H_{max}=\frac{(38.67)^2sin^2(77.15\degree)}{(2)(10)}=71.07\text{ m} \end{gathered}[/tex]c. Now time of flight of the projectile is given by,
[tex]t_{flight}=\frac{2usin\theta}{g}=\frac{(2)(38.67)sin(77.15\degree)}{(10)}=7.54\text{ s}[/tex]this is the time at which the projectile strikes the ground, and so its velocity then would be,
[tex]\begin{gathered} \vec{v}(7.54s)=8.6\text{ }\hat{i}+(37.7-10\times7.54)\hat{j} \\ \Rightarrow\vec{v}(7.54s)=8.6\text{ }\hat{i}-37.7\text{ }\hat{j} \end{gathered}[/tex]As expected the vertical component of velocity has just been flipped, and therefore its speed will be same as its initial speed i.e.
[tex]v_{strikes\text{ }ground}=u=38.67\text{ m/s}[/tex]d. Below the horizontal, let the angle be φ, then
[tex]tan\varphi=-\frac{v_y}{v_x}[/tex]extra -ve sign since its below the horizontal, when the projectile strikes the ground,
[tex]\begin{gathered} v_y=-37.7\text{ m/s} \\ v_x=8.6\text{ m/s} \\ \Rightarrow tan\varphi=-\frac{(-37.7)}{8.6}=\frac{37.7}{8.6}=tan\theta \\ \Rightarrow\varphi=\theta \\ \Rightarrow\varphi=77.15\degree \end{gathered}[/tex]Result: a. 64.85 m, b. 71.05 m, c. 38.67 m/s, d. 77.15°.Consider two homogeneous bodies of the same volume is it necessary for them to have same mass why
No, two homogeneous bodies of the same volume do have not the same mass.
How do two bodies of the same volume have the different same mass?
No, two objects don't need to have same volume to have the same mass because the density of the objects they are made up of can be different from one another. Mass and volume are independent, two objects with the same volume can have different amounts of masses. So that's why the objects can have different densities. Density is defined as the mass of the substance divided by its volume.
Density = mass /volume
Mass is defined as the amount of matter that is present in an object, while on the other hand, volume is the space occupied by an object. For example, a bowling ball and a basketball have the same volume but the bowling ball has more mass.
So we can conclude that two homogeneous bodies of the same volume do have not the same mass because of the difference in their volume.
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The position of a particle is given by the vector < 3.3t2 , 3.2 , - 2.4t >. What is the magnitude of the acceleration at t=0?
The position of a particle is given by the vector < 3.3t² , 3.2 , - 2.4t >. The magnitude of the acceleration at t=0 is 3.3 m/sec².
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.
Given in the question The position of a particle is given by the vector
< 3.3t² , 3.2 , - 2.4t >. The magnitude of the acceleration at t=0.
Differentiating position vector for two times we get acceleration vector,
so acceleration is 3.3 m/sec²
.The position of a particle is given by the vector < 3.3t² , 3.2 , - 2.4t >. The magnitude of the acceleration at t=0 is 3.3 m/sec²
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Change 72km/h into m/s
Answer:
20 m/s
Explanation:
72 km/hr * 1 hr/ 3600sec * 1000m/km = 20 m/s
A ball is thrown up in the air, it reaches its maximum height and falls back. It’s acceleration is:A - same up and down but zero at the topB- less on the way up than on the way downC- the same ally all points of motionD- less on the way down than on the way up
When the ball is thrown up in the air, it is called a Freefall motion.
In freefall motion, the acceleration remains constant at all times because the acceleration due to gravity is always constant.
The acceleration due to gravity is 9.81 m/s^2
The ball has zero velocity (not acceleration) at the maximum height.
Therefore, the correct answer is option C
"Its acceleration is the same at all points of motion".
A 1,116 kg space vehicle is traveling along a straight-line at a constant speed of 800 m/s. What is the magnitude of the net force on the space vehicle?
The magnitude of the net force on this space vehicle moving at constant speed is zero (0) Newton.
What is a net force?A net force can be defined as the vector sum of all the forces that are acting on a physical object or body. This ultimately implies that, a net force is a single force that substitutes the effect of all the forces acting on a physical object or body.
Generally speaking, the amount of net force that keeps a physical object or body moving at constant speed and in the same direction is always equal to zero (0) Newton in accordance with Newton's First Law of Motion.
In this context, we can reasonably infer and logically deduce that the net force on a physical object or body would be equal to zero (0) Newton when it moves at constant speed.
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Biggs and Smalls play tug-of-war with a 12-meter rope in P.E. class. Both are in their socks on a waxed gym floor, which is nearly frictionless. After a brief time, they meet. If Biggs has twice the mass of Smalls, how far has Biggs moved?
After a brief time, Biggs and smalls would meet at 4m from Biggs' side if the rope is 12 meter long and the floor is frictionless.
[tex]X_{cm}[/tex] = ( m1 x1 + m2 x2 ) / ( m1 + m2 )
[tex]X_{cm}[/tex] = Center of mass
After a brief time, Biggs and smalls would meet at the point of centre of mass. Let Biggs be 1 and smalls be 2 and the distances be measured from Biggs' side assuming that Biggs is standing on the left side.
m1 = 2 * m2
x1 = 0 m
x2 = 12 m
[tex]X_{cm}[/tex] = [ ( 2 * m2 * 0 ) + ( m2 * 12 ) ] / [ ( 2 * m2 ) + m2 ]
[tex]X_{cm}[/tex] = 12 m2 / 3 m2
[tex]X_{cm}[/tex] = 4 m
Center of mass is the average position of all the objects on a system, weighed according to the mass of individual objects. It is a position relative to an object.
Therefore, after a brief time, Biggs and smalls would meet at 4 m from Biggs' side.
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List three ways that you can use water in each of its states.
What is the first step in scientific method?
A) Gather information
B) state the problem
C) form a hypothesis
D) perform the experiment
The first step in scientific method is to state the problem. That is option B.
What are scientific methods?The scientific methods are those methods that are used by scientists and researchers to discover answers to specific questions that are being asked and it involves the use of scientific steps and guidelines.
These steps include the following:
State the problem: This is the first step where the problem to be investigated is being defined.Form a hypothesis: This is the step that involves the making of predictions based on the research.Gather information: The data that is related to the problem are being collected.Perform the experiment: Analysis are being carried out on the collected data.Learn more about scientific methods here:
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Light normally travels in straight lines. What term is used to describe the bending of light when it crosses from one medium to another?
The term refraction is used to describe the bending of light when it crosses from one medium to another. The bending of light happens due to the different densities of the two substances.
Can you please explain to me how a steam engine is a example of the 2nd law of thermodynamics
To find:
How a steam engine is an example of the second law of thermodynamics?
Explanation:
The second law of thermodynamics has more than one statement.
Kelvin-Plank statement:
No process is possible whose sole result is to convert the heat absorbed completely into useful work.
Clausius statement:
No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.
Both of the statements are equivalent.
A steam engine takes the heat from a hot reservoir and converts some parts of it into useful work and the rest of the heat absorbed will be sent to a cold reservoir. Thus the steam engine is an example of the 2nd law of thermodynamics.
Look at the diagram below.
Encoding technique - Decoding technique
"Written language
Pictures
Billboards
Advertisements
E-mails
Reading
Viewing
Interpreting
The communication process, which involves understanding parts and meanings, consists of eight key elements: Source Message Channel Receiver Feedback Environment Context, and Interference
Common examples of character encoding systems are Morse Code, Bodo Code, American Standard Code ASCII for Information Interchange, and Unicode. The purpose of encoding is to transform data so that it can be properly and safely processed by another type of binary system data transmitted via e-mail or special characters that appear on web pages.
The goal is not to keep information secret, but to make it available for proper consumption. Coding is the process of transforming data into the formats required for various information processing needs, such as compiling and running programs. Coding uses different patterns of voltage or current levels to represent the 1s and 0s of a digital signal on a transmission line. Common types of line coding are unipolar, polar, bipolar, and Manchester.
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An ocean fishing boat is drifting just above a school oftuna on a foggy day. Without warning, an engine backfireoccurs on another boat at a distance of d = 1.55 km(Figure 1). The speed of sound in air is 343 m/s, and insea water is 1560 m/s.How much time elapsed before the backfire is heard by the fishermen? Assume that the fishermen hears the sound that travels only through air.
For the fisher man use speed of sound in air as v = 343
d= vt
t= d/v
where:
t= time
d= distance = 1.55 km = 1550 m
v= speed= 343 m/s
t= 1550m / 343 m/s = 4.52 s
2. Name the force or forces that cause these objects to
experience changes in motion. Neglect air resistance. Rw
(a) A ball falls toward the ground.
(b) A person accelerates up in an elevator.
(c) A car gradually slows down while approaching a red
light.
a) A ball falls toward the ground:
The force of gravity is acting on the falling ball. On the surface of the earth, the direction of this force is always downward, towards the ground. It pulls on all objects with mass.
b)A person accelerates up in an elevator:
On the person in the elevator, there are two forces:
1) Normal Force upward, exerted by the floor on the person perpendicular to the surface (“normal” means perpendicular).
2) Gravitational Force downward, exerted by the Earth on the person. This is also called the person’s “weight”.
c) A car gradually slows down while approaching a red light.:
Friction may slow the car down as it moves along the road, but it also is the force that enables the car to move forward at all. It is the friction force that keeps the tires from sliding on the road. By the same token, it is friction that makes the car come to a stop when the brakes are applied.
What is Force ?
An object's push or pull is seen as exerting a force. The interaction of the objects produces push and pull. You can also use words like stretch and squeeze to describe force. The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force. A spring balance can be used to calculate the Force. Newton is the SI unit of force (N).
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A football is kicked with a velocity of 32.0m/s and at an angle of 42 degrees it takes 4.40s. How high does it travel?
23.4 m
Explanation
to solve this we need to use the expression:
[tex]h=v\sin (\theta)t-\frac{gt^2}{2}[/tex]where
[tex]\begin{gathered} h\text{ is the heigth} \\ v\text{ is initial velocity} \\ \theta\text{ is the angle} \\ t\text{ is the time} \end{gathered}[/tex]then, let
[tex]\begin{gathered} v=32\text{ m/s} \\ \theta=42\text{ \degree} \\ t=4.4\text{ s} \\ g=9.8\text{ }\frac{m}{s^2} \end{gathered}[/tex]now, replace
[tex]\begin{gathered} h=v\sin (\theta)t-\frac{gt^2}{2} \\ h=32\sin (42)(4.40)-\frac{(9.8)(4.4^2)}{2} \\ h=94.21-94.864 \\ h=-0.654 \end{gathered}[/tex]let's check the time of flight
[tex]\begin{gathered} t=\frac{2v\sin\theta}{g} \\ t=\frac{2\cdot32\cdot\sin(42)}{9.8} \\ t=4.36 \end{gathered}[/tex]it means after 4.36 the ball is on the ground.
Step 2
so, to find the maximum heigth we need to use the expression
[tex]\begin{gathered} y_{\max }=\frac{v^2\sin ^2\theta}{2g} \\ \text{replace} \\ y_{\max }=\frac{(32)^2\sin ^2(42)}{2(9.8)} \\ y_{\max }=\frac{1024\cdot0.4477}{2(9.8)} \\ y_{\max }=\frac{458.48}{19.6} \\ y_{\max }=23.39 \end{gathered}[/tex]therefore, the answer is
23.4 m
I hope this helps you
