A golf ball is dropped from rest from a height of 9.00 m. It hits the pavement,

Q: A golf ball is dropped from rest from a height of 9.00 m. It hits the pavement,

1. time to drop from 9 m , yf - yi = vi t + a t^2/ 2 0 - 9 = 0 - 9.8 t^2 /2 t1 = 1.355 sec after its max height is 5.2 m, vf^2 - vi^2 = 2 a (yf - yi) 0^2 - (v0)^2 = 2(-9.8)(5.2 - 0) v0 = 10.1 m/s ....this is bound back speed. So yi = 0 and yf = 1 m putting, 1 - 0 = 10.1 t2^2 - 4.9 t2^2 4.9 t2^2 - 10.1 t2 + 1 = 0 t2 = 1.957 sec total time = t1 + t2 = 3.31 sec

ID: 1884960 • Letter: A

Question

A golf ball is dropped from rest from a height of 9.00 m. It hits the pavement, then bounces back up, rising just 5.20 m before falling back down again. A boy then catches the ball when it is 1.00 m above the pavement.

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s 121 Practice problems Wednesday September 26, 2011s Question A golf ball is dropped from rest from a height of 9.00 m. It hits the pavement, then bounces back up, rising just 5.20 m before falling back down again. A boy then catches the ball when it is 1.00 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch. Sa /ty h2 a 00

Explanation / Answer

time to drop from 9 m ,

yf - yi = vi t + a t^2/ 2

0 - 9 = 0 - 9.8 t^2 /2

t1 = 1.355 sec

after its max height is 5.2 m,

vf^2 - vi^2 = 2 a (yf - yi)

0^2 - (v0)^2 = 2(-9.8)(5.2 - 0)

v0 = 10.1 m/s ....this is bound back speed.

So yi = 0 and yf = 1 m

putting, 1 - 0 = 10.1 t2^2 - 4.9 t2^2

4.9 t2^2 - 10.1 t2 + 1 = 0

t2 = 1.957 sec

total time = t1 + t2 = 3.31 sec

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A golf ball is dropped from rest from a height of 9.00 m. It hits the pavement,

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