Question

You throw ball of mass 92 g upwards with an initial speed of 17 m/s. The work done by the force of gravity on the ball during its trip to its maximum height is

Answer 1

Solution:-

Initial velocity of the ball, u = 17 m/s
Final velocity of the ball, v = 0 (At maximum height, the velocity of the ball becomes zero)
Acceleration, a =g=9.8 m/s2
From third equation of motion, height (s) can be calculated as:
v2−u2 = 2gs
s = (v2−u2)/2g = ((0)2−(17)2)/2(−9.8) = 14.75 m

The work done by the force of gravity on the ball during its trip to its maximum height is given by

U = mgh = 92*10-3 * 9.8*14.75 = 13.2986 J