Question -
Answer -
Initial velocity ofthe ball, u = 49 m/s
Case : I
The boy throws theball upwards when the lift is stationary. The vertically upward direction istaken as the positive direction. The displacement of the ball is zero.
Considering theequation of motion
s=ut + (1/2)at2
0 = (49)t + (1/2)(-9.8)t2
t = (49 x 2)/9.8 =98/9.8 = 10 sec
Case : II
As the lift startsmoving with a speed of 5 m/s, the initial speed of the ball will be 49 m/s + 5m/s = 54 m/s
The displacement ofthe ball will be s = 5t’
Therefore, the timetaken can be calculated using the formula
s = ut + (1/2) at2
5t’ = (54) t’ +(1/2)(-9.8) t’2
t’ = 2(54 – 5)/9.8 =10 sec
The time taken willremain the same in both the cases.