Question

A person is going up in a lift with constant acceleration 2 m/s2. When the lift speed is 10 m/s, he drops a coin from a height of 2 m. Coin will hit the lift floor at time :(Take g = 10m/s2).​

Answer 1

Answer:

The answer is 0.63 s

Explanation:

Using the second equation of motion:

S = vt + 1/2 at2

Here S = h

h = vt + 1/2 gt2

Given data:

Acceleration, a = 2 m/s2  

Speed, v = 10 m/s2

Height, h = 2 m  

The coin is dropped so a = +g = 10 m/s2 and v = o

h = vt + 1/2 gt2

2 = o x t + 1/2 x 10 x t2

2 = 0 + 5 x t2

2 = 5 x t2

t2 = 2 / 5 =0.4

Taking square root  

t = 0.63 s