Question

The maximum height reached by the projectile is $4m$. The horizontal range is $12m$. Velocity of projection in $m/s$ is ($g$ - acceleration due to gravity)

• A. $5\sqrt{\frac{g}{2}}$
• B. $3\sqrt{\frac{g}{2}}$
• C. $\frac{1}{3}\sqrt{\frac{g}{2}}$
• D. $\frac{1}{5}\sqrt{\frac{g}{2}}$

Answer 1

The correct option is A $5\sqrt{\frac{g}{2}}$

in a projectile projected at angle $\theta$ maximum height and range is given by:

$H_{max}=4=\frac{u^2{\sin }^2\theta }{2g}-\left(1\right)$
$R=12=\frac{u^2\sin 2\theta }{g}-\left(2\right)$

Dividing equation (2) by equation (1) :

$\Rightarrow 3=\frac{\sin 2\theta }{{\sin }^2\theta }\times 2$

$\frac{3}{2}=\frac{2\sin \theta \cos \theta }{{\sin }^2\theta }$

$\Rightarrow \tan \theta =\frac{4}{3}\Rightarrow \sin \theta =\frac{4}{5}.$

Using equation (1);
$\Rightarrow 4=\frac{u^2}{2g}\times {\left(\frac{4}{5}\right)}^2$

$\Rightarrow u^2=\frac{25g}{2}$
$\therefore u=5\sqrt{\frac{g}{2}}$