Projectile Motion Analysis: Golf Ball Speed, Distance, and Height

[cahp19]

Homework Statement

A golf ball is struck at ground level. The speed of the golf ball as a function of time is shown in the figure below, where t = 0 at the instant the ball is struck. The graph is marked in increments of 0.55 s along the time axis, and vmin = 18.92 m/s and vmax = 32.93 m/s. (Values in figure do not necessarily match values in problem).

This graph is then provided:

a) How far does the golf ball travel horizontally before returning to ground level?
b) What is the maximum height above the ground level attained by the ball?

Homework Equations

I'm not sure, because each applicable equation I find has the need for a theta in it.

The Attempt at a Solution

I can't even find a working equation.

Thanks a lot!

 

[gneill]

What values can you pick out from the given information? List them.
(Hint: some may require looking for points of interest on the graph and measuring/deducing values).

 

[cahp19]

What values can you pick out from the given information? List them.
(Hint: some may require looking for points of interest on the graph and measuring/deducing values).

The points that I am able to get from the graph are:
t(0)= 32.93 m/s
t(2.75)= 18.92 m/s
t(5.5)= 32.93 m/s

 

[zgozvrm]

You also know the total time the ball is in flight.

Plus what does the value of Vmin=18.92 m/s tell you about the vertical and/or horizontal velocity at that time?

 

[rwisz]

Hello,

I would first like to clarify that without knowing the specific values and units used in the graph and problem, it is difficult to provide a precise answer. However, I can provide some general information and equations that may help in solving this problem.

Firstly, the motion of the golf ball can be described using the equations of projectile motion, which take into account the initial velocity, angle of launch, and acceleration due to gravity. The initial velocity in this case would be the minimum speed of 18.92 m/s and the maximum speed of 32.93 m/s, while the angle of launch can be assumed to be 45 degrees for simplicity.

To answer the first question, we can use the equation for horizontal displacement (range) of a projectile, which is given by:
R = (v0^2/g) * sin(2θ)
Where v0 is the initial velocity, g is the acceleration due to gravity, and θ is the angle of launch. In this case, since the angle is 45 degrees, we can simplify the equation to:
R = (v0^2/g)

Using the minimum velocity of 18.92 m/s, we can calculate the horizontal displacement as:
R = (18.92^2/9.8) = 36.13 m

For the second question, we can use the equation for maximum height of a projectile, which is given by:
hmax = (v0^2/2g) * sin^2(θ)
Again, since the angle is 45 degrees, we can simplify the equation to:
hmax = (v0^2/2g)

Using the maximum velocity of 32.93 m/s, we can calculate the maximum height as:
hmax = (32.93^2/2*9.8) = 57.91 m

In conclusion, the golf ball travels 36.13 meters horizontally before returning to ground level and reaches a maximum height of 57.91 meters above ground level.

I hope this helps in your analysis of the projectile motion of a golf ball. It is important to note that the values and equations used may vary depending on the specific units and values given in the problem. It is always best to double check your calculations and use the appropriate equations for the given scenario.

Best of luck with your studies!