Question

The maximum height reached by projectile is $4\text{m}$. The horizontal range is $12\text{m}$. Velocity of projection in $\left({\text{ms}}^{-1}\right)$ is
($g$ is 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 $5\sqrt{\frac{g}{2}}$

By using relation, $4H=R\tan \theta$, we get
$4\left(4\right)=12\tan \theta$
$\Rightarrow \tan \theta =\frac{4}{3}\Rightarrow \theta ={53}^{\circ }$
$H=4\text{m}=\frac{u^2{\sin }^2\theta }{2g}$
$4=\frac{u^2{\sin }^2{53}^{\circ }}{2g}$
$u^2=8g\times \frac{25}{16}$
$u=5\sqrt{\frac{g}{2}}$