What Is the Height of the Fence Cleared by the Golf Ball?

AI Thread Summary
A golf ball hit at ground level reaches its maximum height after 4.3 seconds and clears a fence 1.2 seconds later, located 309 feet away. The acceleration due to gravity is 32 ft/s², and the calculated height of the fence is 272.8 feet. Participants in the discussion emphasize the importance of using the correct mechanics equations to find both the initial vertical velocity and the maximum height of the ball. There is acknowledgment that the angle of the hit affects the trajectory, but sufficient vertical velocity is necessary for the ball to rise for the given time. The conversation highlights the need for careful calculations and consideration of realistic scenarios in physics problems.
sterlinghubbard
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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.

[edit] Solution is 272.8 [/edit]
 
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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.
 
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.
 
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.
 
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.
 
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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