Question

A particle is drop from tower of height $h$. It covers height $\frac{11h}{36}$ in last second of its journey find height of tower and time taken to cover this height -

• A. $\text{36 m, 6 sec}$
• B. $\text{180 m, 6 sec}$
• C. $\text{180 m, 3 sec}$
• D. $\text{125 m, 5 sec}$

Answer 1

The correct option is B $\text{180 m, 6 sec}$

If t is time to cover height h then $\sqrt{h=\frac{1}{2}9t^2}$

suppose height fall in $t-1$ sec is $h^{\prime }$

it will be $h^{{}^{\prime }}=\frac{1}{2}9(t-1{)}^2=\frac{1}{2}9\times {\left(\sqrt{\frac{2h}{9}-1}\right)}^2$

given that $h-h^{\prime }=\frac{11h}{36}$

$\Rightarrow \frac{1}{2}9t^2-d\frac{1}{2}9(t-1{)}^2=\frac{11h}{36}$

$\Rightarrow t^2-(t-1{)}^2=\frac{22h}{369}$

or $(t+t-1)(t-t+1)=\frac{22h}{369}=\frac{11h}{189}$

$2t-1=\frac{11h}{189}$

Now put $h=\frac{1}{2}9t^2$

we get $2t-1=\frac{11}{189}\times \frac{1}{2}9t^2=\frac{11}{36}{t}^2$

$72t-36=11t^2$ or $11t^2-72t+36=0$

using shridharacharya formula

we get $t=\frac{72\pm \sqrt{{72}^2-4\times 36\times 11}}{2\times 11}$

$[t=68ec]$, $0.5$ sec x

$\Rightarrow [h=\frac{1}{2}9t^2=\frac{1}{2}\times 10\times 36=180m]$.