Question

Stone is dropped from a tower.It covers 25 m in last second of motion.Find height of tower

Answer 1

In one (last) second of motion, we know that the acceleration of the stone would be $g = 10m/s^2$.

So, using $S = ut + \frac{1}{2}at^2$

At time (t - 1): $S_{t - 1} = u(t - 1) + 5(t - 1)^2$

And, at t: $S_{t} = ut + 5t^2$

Considering t = 0 when the stone was dropped, u shall be zero (zero initial velocity):

$S_{t - 1} = 5(t^2 - 2t + 1)$

And, $S_{t} = 5t^2$

It is given that the distance covered in the last second is 25m, so:

$S_{t} - S_{t - 1} = 25m = 5t^2 - (5(t^2 - 2t + 1)) = 10t - 5$

So, $10t = 30$ (ignoring units)

And, $t = 3s$

So, it took the stone 3 seconds to cover the whole height of the tower, while free-falling.

For the height of the tower, we use the motion of equation again:

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

With u = 0, t = 3, a = 10

$S = \frac{1}{2} * 10 * (3)^2 = 5 * 9 = 45$

So, S, the height of the tower = 45m.