Question

an elevator is moving upward with a constant acceleration of 0.2 metre per second square. at some instant when speed of elevator is 10 metre per second the man drops a coin from a height of 20m. the time taken by the coin to reach the floor is​

Answer 1

Answer:

t = 2 sec

Explanation:

An elevator is moving upward with a constant acceleration of 0.2 metre per second square. at some instant when speed of elevator is 10 metre per second the man drops a coin from a height of 20m. the time taken by the coin to reach the floor is​

Speed of Elevator = 10 m/s

Relative speed of coin = 0 m/s

Net acceleration of coin  = 9.8 + 0.2 = 10 m/s²

Distance = 20 m

S = ut + (1/2)at²

=> 20 = 0 + (1/2)(10)t²

=> 20 = 0 + 5t²

=> 4 =  t²

=> t = 2  sec

Answer 2

answer : 2sec

an elevator is moving upward with a constant acceleration, a = 0.2 m/s² . it means elevator is an non-inertial frame of reference. and we know, we cannot apply Newtonian equation in non-inertial frame of reference. so, first of all Let's make it inertial frame of reference. applying a Pseudo force of equal magnitude just opposite of its motion to make it inertial frame.

Pseudo force, F = ma

so, Pseudo acceleration, a =- 0.2m/s² (negative sign shows downward direction)

now, net acceleration acting on coin, A = g + a = -9.8m/s² +(-0.2 m/s² )= -10m/s²

a/c to question, at instant when speed of elevator is 10m/s. then man drop a coin from a height of 20m.

so, initial velocity of coin, u = 0

and acceleration acting on coin, A = 10m/s²

so, using formula, S = ut + 1/2 at²

here, S = 20m, a = A = -10m/s², u = 0,

then, 20 = 0 + 1/2 × 10t²

or, t² = 4 => t = 2sec .