Question

after falling from rest under gravity for 5 seconds ,a ball passes through a pane of glass held horizontally and instantaneously loses half its velocity . if the ball then takes one more second to reach the ground , calculate the height of the glass above the ground (g=10ms^-2)

Answer 1

Let v1 be speed of ball when it reaches the glass

and v2 be the speed of the ball when it reaches the ground.

v1 = g t = 50 m/s

v1 / 2 = 25 = speed of ball after passing thru glass

v2 = v1 /2 + g t2 = 25 + 10 = 35 m/s

2 g h = v2^2 - v1^2 = 35^2 - 25^2 = 600

h = 600 / 20 = 30 m