Question

A particle is dropped from the top of a tower. During its motion it covers $\frac{9}{25}$ part of height of tower the last 1 seconds. Then find the height of tower.

Answer 1

Ball travels $\frac{9}{25}$ of the height (say\space $x$) in the $t$ second (last second)

$\Rightarrow$Hence it travelled $\frac{16}{25}x$ in $(t-1)$ seconds.

and the total distance in $t$ seconds.

The ball is dropped hence initial velocity is zero and gravity is $9.8\frac{m}{s^2}$

Hence distance travelled is

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

$\Rightarrow x=\frac{1}{2}g{t}^2$ ……$\left(1\right)$

$\Rightarrow \frac{16x}{25}=\frac{1}{2}g(t-1{)}^2$ …..$\left(2\right)$

Divide the equations $\left(2\right)$ by $\left(1\right)$

$\frac{16}{25}=\frac{(t-1{)}^2}{t^2}$

$\Rightarrow 16t^2=25(t-1{)}^2$ …..(3)

Which gives $t=5$ seconds (Time cannot be negative)

Hence height $x=\frac{1}{2}\times 9.8\times \left(5^2\right)=122.5m$