How far beyond the fence does the ball strike the ground?

[goaliejoe35]

Homework Statement
A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.6 s after being hit. Then 3.1 s after reaching its maximum height, the ball barely clears a fence that is 87.5 m from where it was hit. Assume the ground is level. (a) What maximum height above ground level is reached by the ball? (b) How high is the fence? (c) How far beyond the fence does the ball strike the ground?

The attempt at a solution
(a) 63.5 m

(b) 16.4 m

(c) ??

I got these answers by using H=.5gt^2

(a) H=.5(9.8)(3.6)^2 = 63.5 m <--max height

(b) H=.5(9.8)(3.1)^2 = 47.1 m
63.5-47.1 = 16.4 m <--height of fence

How do I go about finding part (c) How far beyond the fence does the ball strike the ground?

 

[ace123]

Well you can calculate the time it takes to strike the ground after clearing the fence.

 

[arunbg]

For part c, you need to calculate horizontal distance, and for this you need to use the fact that the fence is at a distance of 87.5m from hit. Can you calculate the horizontal component of velocity?

 

[goaliejoe35]

Would horizontal component for velocity be time? If so would it be 6.7?

 

[ace123]

The horizontal component doesn't change it's constant. So if you have the time it takes to get to the fence and the distance to the fence you can calculate the horizontal component

 

[goaliejoe35]

When you say horizontal component what do you mean?

 

[ace123]

I mean the velocity in the horizontal direction. The baseball has a vertical and horizontal velocity. The vertical velocity is changing the horizontal is not changing. Understand?

 

[goaliejoe35]

so like V= 87.5/6.7 = 13.1 m/s

 

[ace123]

Yes. But you must now find the distance traveled beyond the fence

 

[goaliejoe35]

Ok so would I set up an equation like this x= vt and if that's true how would I know what t equals?

 

[ace123]

Well you know the time it takes to reach the maximum height and the time it takes to reach the fence. So what would you do to find the time to reach the ground beyond the fence?

Hint: what is the total time?

 

[goaliejoe35]

3.6s + 3.1s = 6.7 s to reach the fence then I'm lost. How would I calculate the time to reach the ground? I know that the velocity when it hits the ground is 0 correct?

 

[ace123]

Okay think of it another way. When you throw a baseball up in the air it takes 3 seconds to reach the maximum height then it start coming back down to you. How long would that take?

 

[goaliejoe35]

3 seconds?

 

[ace123]

Exactly. Can you apply this to your current problem?

 

[goaliejoe35]

use the equation v=xt and solve for x when v is 13.1 m/s and t is 0.5s?

 

[arunbg]

You simply have to find the distance it takes to reach max height and double it.

 

[ace123]

You simply have to find the distance it takes to reach max height and double it.

You mean the time.

 

[ace123]

V doesn't equal to xt. V=x/t

 

[goaliejoe35]

i meant x=vt

and solve for x like this

x=13.1(7.2)
x=94.32 m

94.32-87.5= 6.82m

is that right?

 

[ace123]

Yea I figured but I just wanted to make sure

 

[arunbg]

You mean the time.

No, distance covered to reach the highest point is same as distance covered when coming down from the highest point, since horizontal velocity is constant. Time taken also are the same.

 

[ace123]

No, distance covered to reach the highest point is same as distance covered when coming down from the highest point, since horizontal velocity is constant. Time taken also are the same.

I was correcting your english. The distance it takes to reach the maximum height. See what I mean.

 

[ace123]

Yes that is correct. I wouldn't have rounded the velocity up to 13.1. Instead I would have left it till the end of the problem and had my final answer to the correct sig figs because my answer is 6.53 when you don't round.

 

[arunbg]

Oh sorry, bad construction. I meant horizontal distance it takes to reach maximum height of course. Thanks ace.

 

[ace123]

No harm. Thanks for helping me on the other problem.