A golf ball is dropped from rest from a height of 8.80 m. It hits the pavement,

Question

A golf ball is dropped from rest from a height of 8.80 m. It hits the pavement, then bounces back up, rising just 5.80 m before falling back down again. A boy then catches the ball when it is 1.20 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.

Explanation / Answer

the time for the ball reached to the pavement and then bounces: x g*x^2/2= 8.8 =>x=1.34 s the time for the ball bounces back to the height of 5.8 m and catched 5.8=g*y^2/2 and (5.8-1.2)=g*z^2/2 and z=0.968 y=1.089 => y+z= 2.057 s => total time is:t=x+y+z=3.37 s

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