Question

An elevator, in which a man is standing, is moving upward with a constant acceleration of 1m/s^2. At some instant when speed of elevator is 10 m/s, the man drops a coin from a height of 2 m. Find the time taken by the coin to reach the floor. (g 9.8m/s')​

Answer 1

Answer:

s=

[(9.8+1)t^2]/2

Answer 2

Answer:

1/6 and 2¹/54

Explanation:

s=ut+1/2at²

here , relative acceleration=9.8+1=10.8m/s²

u=10m/s

therefore, equation

2=10t+1/2×10.8×t²

solving this,

t =1/6 and 2 ¹/54(since the equation is quadratic)