A golf ball is dropped from rest from a height of 9.5 m. It hits the pavement, then bounces back up, rising just 5.7 m before falling back down again. A boy 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.

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.

Answered by avtrisal | 2024-06-10

Answer: 3.4s ;

Answered by Edufirst | 2024-06-11

The total time the golf ball is in the air, from drop to catch, is approximately 3.4 seconds. This includes the time taken to fall, rise, and then fall again before being caught. Each stage of motion was analyzed separately to calculate the overall time.
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Answered by Edufirst | 2024-12-23