Physics word problem

Question

A golf ball is dropped from rest from a height of 8.70 m. It hits the pavement, then bounces back up, rising just 5.60 m before falling back down again. A boy then catches the ball when it is 1.40 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.

Answer 1

Understanding the Problem

We're given a ball dropped from a height, bouncing, and then being caught. We need to find the total time the ball is in the air.

Breaking Down the Problem

We can break this down into three parts:

1. Fall from initial height to the pavement.
2. Rise after the bounce.
3. Fall from the bounce height to the catch point.



Solution

Part 1: Fall from initial height to the pavement

1. Equation: h=21​gt2
2. Solving for t: t=g2h​​
3. Time: t1​=9.8m/s22×8.70m​​≈1.33seconds

Part 2: Rise after the bounce

1. Equation: h=21​gt2
2. Solving for t: t=g2h​​
3. Time: t2​=9.8m/s22×5.60m​​≈1.07seconds
4. Total time for rise and fall: tbounce​=2t2​≈2.14seconds

Part 3: Fall from bounce height to catch point

1. Distance fallen: h=5.60m−1.40m=4.20m
2. Time: t3​=9.8m/s22×4.20m​​≈0.93seconds

Total Time

1. ttotal​=t1​+tbounce​+t3​≈1.33+2.14+0.93≈4.40seconds

Therefore, the ball is in the air for approximately 4.40 seconds.

Answer 2

Hi Tarah,

It's good to break up this big problem into smaller (easier) problems. I'd suggest finding the time it takes to

1. Drop the initial 8.70 m
2. Go up the 5.60 m
3. Come back down the distance 5.60 m - 1.40 m

Then add up those three times. For each of these three "segments" you can use the equations learned in class for motion in 1 dimension with constant acceleration. In this problem you have what's called free fall, wherein the acceleration is only due to gravity and is 9.8 m/s^2, pointing down -- that usually means a = -9.8 m/s^2, which assumes that you have chosen the y-direction to point upward.

I hope that helps!