Question

The maximum height reached by projectile is $4m$. The horizontal range is $12m$. Velocity of projection (in $m{s}^{-1}$) is
($g$ is acceleration due to gravity)

• A. $5\sqrt{\frac{g}{2}}$
• B. $3\sqrt{\frac{g}{2}}$
• C. $\frac{1}{2}\sqrt{\frac{g}{2}}$
• 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
$4\left(4\right)=12\tan \theta \Rightarrow \tan \theta =\frac{4}{3}\Rightarrow \theta ={53}^{\circ }H=4m=\frac{u^2{\sin }^2\theta }{2g}4=\frac{u^2{\sin }^2{53}^{\circ }}{2g}{u}^2=8g\times \frac{25}{16}u=5\sqrt{\frac{g}{2}}m/s$