Question

The maximum height attained by a projectile is increased by $5\%$. Keeping the angle of projection constant, what is the percentage increase in horizontal range?

• A. $5$%
• B. $10$%
• C. $15$%
• D. $20$%

Answer 1

The correct option is A $5$%

If $h$ is the maximum height attained by the projectile,
then $h=\frac{u^2{\sin }^2\theta }{2g}$ or $R=\frac{u^2\sin 2\theta }{g}$

$\frac{R}{h}=\frac{2\sin \theta \cos \theta }{\left({\sin }^2\theta \right)/2}=4\cot h\theta$

$\therefore \frac{△R}{R}=\frac{△h}{h}$

Percentage increase in $R=$ percentage increase in $h=5\%$.

$\therefore \frac{\Delta H}{H}=\frac{5}{100}=0.05$

$\therefore \frac{2\Delta u}{u}=0.05$

As, $R=\frac{u^{2}sin2\theta }{g}$

Therefore, $\Delta R=\frac{2u\Delta u}{g}\times sin2\theta$

and $\frac{\Delta R}{R}=\frac{\frac{2u\Delta u}{g}\times sin2\theta }{\frac{u^{2}sin2\theta }{g}}$=$\frac{2\Delta u}{u}=0.05$

$\therefore$% increase in horizontal range $=\frac{\Delta R}{R}\times 100=0.05\times 100=5$%