Question

The maximum height attained by a projectile is increased by $10\%$. Keeping the angle of projection constant, what is the percentage increase in the time of flight?

• A. $5\%$
• B. $20\%$
• C. $10\%$
• D. $40\%$

Answer 1

The correct option is A $5\%$

Let the initial velocity of projection be $‘u^{\prime }$, the initial angle of projection $‘{\theta }^{\prime }$ and time of flight be $‘t^{\prime }.$

Initial max. height is, $h=\frac{\left(u^{2}si{n}^2\theta \right)}{2g}$

Time of flight,$t=\frac{2usin\theta }{g}$

$\Rightarrow t^2=\frac{4u^{2}si{n}^2\theta }{g^2}$

$\Rightarrow t^2=\left(\frac{8}{g}\right)\left[\frac{\left(u^{2}si{n}^2\theta \right)}{2g}\right]=\left(\frac{8}{g}\right)h$

Therefore,

$\frac{T^2}{t^2}=\frac{H}{h}$

Now, new height, $H=h+\left(\frac{10}{100}\right)h=\left(\frac{11}{10}\right)h$

$\Rightarrow \frac{H}{h}=\frac{11}{10}$

And,

$\frac{T^2}{t^2}=\frac{11}{10}$

$\Rightarrow \frac{T}{t}=1.05$

$\Rightarrow \frac{T}{t}–1=0.05$

$\Rightarrow (T-t)t=0.05$

$\Rightarrow (T-t)t\times 100=5$%

Thus, the % increase in time of flight is $5$%.