Question

A particle is dropped from the top of a tower. The distance covered by it in the last one second is equal to that covered by it in the first three seconds. Find the height of the tower.

Answer 1

The distance covered in first 3 second

$s=\frac{1}{2}g{t}^2$

$=12\times 10\times 3^2$

$=45$

If the ball takes 'n' second to fall to the ground. The distance covered in $n^{th}$ second

$s_n=u+\frac{g}{2}(2n-1)$

$=0+\frac{10}{2}(2n-1)$

$=10n-5$

Therefore,

$45=10n-5$

$n=5$

Therefore,

$h=\frac{1}{2}g{t}^2$

$=\frac{1}{2}\times 10\times 25$

$=125m$

Thus the height of the tower is $125m$