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 5 m s¯¹ and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands ?
Case 1: Lift is stationery
Given: The initial speed of the ball is 49 m/s and the life is stationery.
The second equation of motion is given as,
s=ut+12at2
Where s is the displacement of the ball, u is the initial velocity, t is the time taken by the ball and a is the acceleration of the ball in the lift.
In this case, displacement is 0 since the ball comes back to the boy who is standing on the stationary lift.
By substituting the given values in the above equation, we get
0=49×t−12×9.8×t2
Thus, the time taken by the ball to return to the boy's hands when the lift is stationary is 10 s.
Case 2: The lift is moving with a uniform velocity of 5 m/s
The lift is moving with a uniform velocity of 5 m/s in the upward direction which means there is no acceleration in the lift. So in this case also, the relative speed of the ball with respect to the boy remains the same that is 49 m/s. Therefore, the time taken by the ball to return to the boy's hand will same as in case I.
Thus, the time taken by the ball to return to the boy's hands when the lift is stationary is 10 s.
Case 1: Lift is stationery
Given: The initial speed of the ball is 49 m/s and the life is stationery.
The second equation of motion is given as,
s=ut+12at2
Where s is the displacement of the ball, u is the initial velocity, t is the time taken by the ball and a is the acceleration of the ball in the lift.
In this case, displacement is 0 since the ball comes back to the boy who is standing on the stationary lift.
By substituting the given values in the above equation, we get
0=49×t−12×9.8×t2
Thus, the time taken by the ball to return to the boy's hands when the lift is stationary is 10 s.
Case 2: The lift is moving with a uniform velocity of 5 m/s
The lift is moving with a uniform velocity of 5 m/s in the upward direction which means there is no acceleration in the lift. So in this case also, the relative speed of the ball with respect to the boy remains the same that is 49 m/s. Therefore, the time taken by the ball to return to the boy's hand will same as in case I.
Thus, the time taken by the ball to return to the boy's hands when the lift is stationary is 10 s.
