# Homework help.

#### sterlinghubbard

Homework help. (Solved)

A golf ball is hit at ground level. The ball is observed to reach its maximum height above ground level 4.3s after being hit. 1.2s after reaching this maximum height, the ball is observed to barely clear a fence taht is 309 ft from where it was hit. The accelerration of gravity is 32 ft/s^2. How high is the fence? Answer in units of ft.

The horizontal velocity is constant and 56.18 as defined as distance over time. The vertical velocity is variable but I found the instantaneous velocity (137.6) in an attempt to find the initial angle. I am stumped. I keep coming up with values around 200 ft and that doesn't seem possible.

 Solution is 272.8 [/edit]

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#### WhirlwindMonk

Could you show your work? If you show it we could probably point out what you did wrong.

But I assume you are using the basic mechanics equations? The one you probably would want to start with is v-final^2 = v-initial^2 + 2*g*h.

#### OlderDan

Homework Helper
A ball that rises for 4.3 seconds before reaching maximum height is going to go pretty high. In another 1.2 seconds it is not going to fall terribly far. You might feel better about your answer if you calculatied the maximum height of the ball as well as the height of the fence.

#### WhirlwindMonk

OlderDan said:
A ball that rises for 4.3 seconds before reaching maximum height is going to go pretty high. In another 1.2 seconds it is not going to fall terribly far. You might feel better about your answer if you calculatied the maximum height of the ball as well as the height of the fence.
But it is not hit straight up, it is hit at an angle. If it is hit at a low enough angle, it could rise to only 10 feet in that time.

Ignore my previous post, i was wrong. You want v-final = v-initial + g*t. Calculate the inital vertical velocity and then you should be able to go from there. Assuming I didn't mess up again.

#### whozum

WhirlwindMonk said:
But it is not hit straight up, it is hit at an angle. If it is hit at a low enough angle, it could rise to only 10 feet in that time.

Ignore my previous post, i was wrong. You want v-final = v-initial + g*t. Calculate the inital vertical velocity and then you should be able to go from there. Assuming I didn't mess up again.
Regardless of angle, you must have adequate vertical velocity in order to RISE for 4.3 seconds.

#### WhirlwindMonk

whozum said:
Regardless of angle, you must have adequate vertical velocity in order to RISE for 4.3 seconds.
Ah. True. I'm not thinking, it's almost midnight here. Now that you mention it, it will go up a significant amount. sterlinghubbard, one thing you also want to think about is that books don't always have the most realistic answers. I remember a calc problem with a 6.5 foot tall thief running as fast as an olympic sprinter.

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