Question

An elevator car is moving upward with uniform velocity of 2 m/s. When it is 5 m above the ground a stone is thrown upward from the floor of the lift. The stone strikes back the lift 2 s after its projection. (a) Find the velocity of projection of the stone relative to the lift. (6) Find the velocity of the stone relative to the lift when it strikes back. Solr:​

Answer 1

Answer:

Explanation:

an elevator acceleration = 2m/s² (upward)

Velocity = 2m/s

Distance covered in 1 sec = ut + (1/2)at²

= 2*1 + (1/2)2(1)² = 3 m

Let say Ball relative velocity = V

Time 1 sec

acceleration = -g = -10 m/s²

Distance covered by ball & lift should be same to strike back

S = V(1) + (1/2)(-10)* 1²

=>3=V-5

=> V = 8

the velocity of the projection of the ball relative to the lift = 8 m/s.