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:

$\frac{1}{\sqrt{3}}s$

Explanation:

1)  When the speed of lift is 10 m/s , person drops a coin from a height of 2m in left.

Just after drop, speed of coin is 10m/s in upward direction.

Using concepts of relative motion, (with respect to lift )

$S_{rel}=-2m ( \because downward)\\u_{rel}=10-10=0 m/s\\a_{rel}=-10-2=-12m/s^2$

2) Now, Using Newton's Equation in relative.

$S_{rel}=u_{rel}t+\frac{1}{2}at^2\\ \\-2=0*t-\frac{1}{2}*12*t^2\\ \\=>-2=-6t^2\\ \\=>t^2=\frac{1}{3}\\ \\ =>t=\frac{1}{\sqrt{3}}s$

Hence, time taken by coin to hit the floor is

$\boxed{t=\frac{1}{\sqrt{3}}s}$