Question

a girl standing on a stationary lift (open from above) throws a ball upwards with an initial speed 50 m/s. the time taken by the ball to return to her hands is ( take g=10m/s) :
a. 5 s
b. 10 s
c. 15 s
d. 20 s
((please explain also.))

Answer 1

heya mate here is the answer hope it helps

Answer 2

Given: 
⇒ The lift is stationary.
⇒ Initial speed, u = 50m/s
⇒ g = -a  = -10m/s
⇒ Total displacement, s = 0 (since the initial and final positions of the ball are the same)

To find:
⇒ Time taken by the ball, t = ?

Solution: 
⇒ Let us assume the total time taken by the ball in going upward and coming down to the hands of the girl be 't'.
⇒ Using the relation, s = ut + 1/2 at^2, we get

⇒ 0 = 50t + 1/2 (-10) t^2
⇒ -50t = -5t^2
⇒ 50t = 5t^2
⇒ 10t = t^2
⇒ t^2/t = 10
⇒ ∴ t = 10s

⇒ Thus time taken by the ball to return the girl's hand is 10s (Option b).

:)