# A man drops a coin from a height 5 m. The man stands in a lift that is moving upwards with a speed of 30 m/s. When does the coin hit the ground if g=9.8 ms^-2?

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### 2 Answers

The coin is dropped from a height of 5 m from the ground by a man moving upwards in a lift at 30 m/s. Let the time taken by the coin to hit the ground be t.

Use the formula S = u*t + (1/2)*a*t^2 where u is the initial velocity, S is the distance traveled in time t and a is the acceleration. As S = -5, u = 30 and a = -9.8

-5 = 30*t - (1/2)*9.8*t^2

=> 4.9t^2 - 30t - 5 = 0

Solving for t gives a positive and a negative root.

The time taken cannot be negative, eliminating the negative root, the time taken is approximately 6.28 s.

The lift with the man is moving upward at a velocity of 30m/s. When the man drops the coin it will also have this 30m/s velocity component to the upward direction. Lets assume the time that coin fall on to the ground is t;

Using velocity equations;

`uarr S = Ut+1/2*g*t^2`

`-5 = 30*t+1/2*-9.8*t^2`

- Here the S=-5 because we are putting the equation to the upward but the distance 5m was measured downward.
- g is also taken as -9.81 due to the same reason.

`-5+4.9t^2-30t = 0 `

Therefore `t = (30+-sqrt(30^2-4xx4.9xx-5))/(2*4.9)`

Once we solve this equation for its roots you will get;

t = 6.28 s or t = -0.162 s

**Since time cannot be negative, the coin will hit the ground after 6.28 seconds.**