Question

A particle is dropped from the tower. The distance travelled by it in the last second is $15\text{m}$. The height of the tower is
[Take $g=10{\text{m/s}}^2$]

• A. $22.5\text{m}$
• B. $20\text{m}$
• C. $25\text{m}$
• D. $30\text{m}$

Answer 1

The correct option is B $20\text{m}$

Given, $u=0,a=g$
As per the question we have
$S_t-S_{t-1}=15$
$\Rightarrow \frac{1}{2}g{t}^2-\frac{1}{2}g(t-1{)}^2=15$
$\Rightarrow t^2-t^2+2t-1=\frac{30}{10}=3$
$\Rightarrow 2t-1=3\Rightarrow t=2\text{sec}$
So, height of the tower is given as
$h=\frac{1}{2}g{t}^2=\frac{1}{2}\times 10\times 2^2=20\text{m}$