Question

A very broad elevator is going up vertically with a constant acceleraton
when its velocity is 4 m/s a ball is projected from the floor of the lift with a speed of
4 m/s relative to the floor at an elevation of 30°. The time taken by the ball to return the floor is
(g = 10 m/s)
​

Answer 1

Dear Piyush,

● Answer -

t = 0.333 s

◆ Explaination -

# Given -

a = 2 m/s^2

u = 4 m/s

θ = 30°

# Solution -

Acceleration of the ball w.r.t. lift is -

a' = a + g

a' = 2 + 10

a' = 12 m/s^2

Time taken by the ball to return to floor is calculated by -

t = 2usinθ/a'

t = 2 × 4 × sin30° / 12

t = 0.333 s

Therefore, Time taken by the ball to return to floor is 0.333 s.

Hope this is useful. Keep asking.