A golf ball is dropped from rest from a height of 9.5m. It hits the pavement then bounces back up rising just 5.7 m before falling back down again. A boy then catches the ball on the way down when it is 1.20 m above the pavement. Ignoring air resistance, calculate the total amount of time the ball is in the air, from drop to catch.

Question

Physics

Yaka

14 July, 05:20

A golf ball is dropped from rest from a height of 9.5m. It hits the pavement then bounces back up rising just 5.7 m before falling back down again. A boy then catches the ball on the way down when it is 1.20 m above the pavement. Ignoring air resistance, calculate the total amount of time the ball is in the air, from drop to catch.

+3

Answers (1)

Know the Answer?

Meyers · Answer 1

The time the ball takes to fall 9.5 meters is the square root of (19/g), where g is gravitational acceleration.

The time it takes to rise to 5.7 meters is the square root of (11.4/g), for the same value of g.

The time it takes to fall from 5.7 meters to 1.2 is the square root of (9/g).

So the answer is [sqrt (19) + sqrt (11.4) + sqrt (9) ]/sqrt (g). If g=10, the answer is 3.39 seconds; if g=9.8, the answer is 3.43 seconds.