Question

A coin has been dropped from a height of 186.39m,The coins speed is increasing every 0.7 seconds by 4.5 m/s and the speed stops increasing at 50 m/s how much time does it take for the coin to fall down? (Initial Speed 2 m/s)​

Answer 1

Given,

• A coin dropped from height 186.39m

• Coins speed increase every 0.7 seconds by 4.5m/s

• Speed stops increasing after it reaches 50m/s

• Initial speed of coin is 2m/s

To find,

Time taken by the coin to stop.

Solution,

To put the problem in simple words, we have

A coin dropped from some height with constant acceleration which acts upon un till the velocity of coin becomes 50m/s.

Calculating the acceleration = $\frac{change in velocity}{time}$

                                               $= \frac{4.5}{0.7} m/s^{2}\\= \frac{45}{7} m/s^{2}$

Now, it is possible the coin reaches the ground even before reaching the maximum velocity that is 50m/s.

Using formula,

                       $2as = v^{2} - u^{2}\\2*\frac{45}{7}*s = 50^{2} - 4^{2}\\s = 194.133m$

This shows that, the coin reaches the ground before gaining the speed of 50m/s

Using formula,

                $s = ut + \frac{1}{2} at^{2}$

We get a quadratic equation in terms of time.

                $186.39 = 2t + \frac{1}{2}*6.428t^{2}\\\\\frac{1}{2}*6.428t^{2} + 2t - 186.39 = 0$

$7.31$ and $-7.93$ are the roots of this equation. But as the time cannot be negative we will just consider $7.31$ as our answer.

Therefore time required by the coin is $7.31sec$.