Question

A lift whose cage is 3 m high is moving up with an acceleration of 2 m/s?. A piece of stone
is dropped from the top of the cage of the lift when its velocity is 8 m/s. If g = 10 m/s', then
the stone will reach the floor of the lift after

Answer 1

Answer:

0.305 seconds

Explanation:

Given :

• Height of the cage = 3 metres

• Acceleration of the lift = 2 m/s²

• Initial velocity of the stone = 8 m/s

• Gravitational force = g = 10 m/s²

To find :

• Time taken for the stone to reach the ground

The initial velocity will be 8 m/s as the velocity will be transferred

Acceleration will be = (10+2) = 12 m/s²

Height = 3 metres

Using second equation of motion :

S = ut+½at²

3 = 8×t+½×12×t²

3=8t+6t²

6t²+8t-3=0

Solving with quadratic equation formula,

t = 0.305 seconds

The time taken is equal to 0.305 seconds

Answer 2

Given:-

• Height of cage =3m

• Acceleration of cage =2m/s²

• Acceleration due to gravity =10m/s²

• ∴Net acceleration of stone =10+2=12m/s²

• Initial velocity =0m/s

To Find:-

• Time taken by stone to reach the floor.

Solution :-

By using 2nd equation of motion

➦ S =ut+1/2at²

➭ 3 = 0×t +1/2×12 × t²

➭ 3 = 6 ×t²

➭ t² = 3/6

➭ t = 1/√2

➭ t = 0.705 s

∴The time taken by stone to reach the floor

is 0.705 s.