Question

An elevator, in which a man is standing moving upward with a constant

speed of 10m/sec2. If a man drops a coin from a height of 2.5m. Find the time
taken by it to reach the oor of the elevation? g=9.8m/sec2

Answer 1

Lift frame :

Initial speed of the coin u=0 m/s

Acceleration a=9.8 m/s

2

Initial height of coin from the floor of elevator h=2.45 m

Time taken by coin to hit the floor T=

g

2H

⟹ T=

9.8

2×2.45

=

2

1

s

Answer 2

We are Given:

velocity of the lift (u₁) = 10 m/s

initial velocity of the coin (u₂) = 10 m/s

(the coin is also moving with the Elevator)

acceleration of the coin (a₂) = -9.8 m/s²

acceleration of the Elevator (a₁) = 0 m/s

Distance covered by the coin (s) = -2.5 m

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Relative Velocity and Acceleration:

Relative velocity:

We will find the velocity and acceleration of the coin with respect to the lift since we are monitoring the motion of the coin

velocity of the coin with respect to the elevator (u₂₁) = u₂ - u₁

u₂₁ = 10 - 10

u₂₁ = 0m/s

Relative Acceleration:

acceleration of the coin with respect to the elevator (a₂₁) = a₂ - a₁

a₂₁ = -9.8 - 0

a₂₁ = -9.8

__________________________________________________________

Solving for the Time taken:

From the second equation of motion, we know that:

s= ut + 1/2at₂

we can rewrite this equation in terms of relative motion:

s = u₂₁(t) + 1/2(a₁₂)t²

Notice that the time and the displacement are not relative, that's because displacement and time will remain the same no matter the frame of reference

replacing the known values in the equation:

-2.5 = (0)(t) + 1/2 (-9.8)(t²)

-2.5 = -4.9(t²)

dividing both sides by -4.9

t² = -2.5 / -4.9

t² = 25/49

t² = (5)² / (7)²

taking the square root of both the sides

t = 5/7 OR 0.71 seconds (approx)

Therefore, the coin will reach the floor of the Elevator in 0.71 seconds