Question

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s⁻¹. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of m s⁻¹ and the boy again throws the ball with the maximum speed he can, how long does the ball take to return to his hands?

Answer 1

Hii dear,

# Answer- 10 s in both cases

# Given-
u = 49 m/s
a = –g = –9.8 m/s^2

# Solution-
A) Case I- Lift is stationary-
- Final velocity of the ball becomes zero at the highest point.
- From first equation of motion, time of ascent is given as:
v = u + at
t = (v–u)/a
t = (0-49)/-9.8
t = 5 s
- Also time of ascent = time of descent.
- Hence, the total time taken by the ball to return to the boy’s hand = 5 + 5 = 10 s.

B) Case II: Lift is moving with velocity 5m/s -
- In this case, the relative velocity of the ball with respect to the boy remains the same i.e., 49 m/s.
- Therefore, in this case also, the ball will return back to the boy’s hand after 10 s.

Hope you got it...

Answer 2

Given :

Initial speed=u=49 m/s

Final velocity at maximum height=v=0 m/s

A= -9..8 m/s 2

T=?

From first equation of motion :

V=u+at

0=49 – 9.8 x t

T=49/9.8

=490/98  

= 5 sec

So time of ascent is 5 seconds and time of descent is also 5 second.

So total time= t= time of ascent + time of descent

= 5s+5 s

=10 sec

b) The uniform velocity of lift does not change the relative motion of ball and lift. So ball would take same time as it would take after 10 sec