Question

The total speed of a projectile at its greatest height is $\sqrt{\frac{6}{7}}$ of its speed when it is at half of its greatest height . The angle of projection will

• A. 40
• B. 45
• C. 30
• D. 50

Answer 1

The correct option is A 40

Let greatest height be ${}^{\prime }{H}^{\prime }$

then$,$

$H=\frac{u^2}{2g}$

at half of its greatest height let

$h=\frac{v^2{\sin }^2\theta }{2g}$

Now$,$ $h=\frac{H}{2}$ $\left(given\right)$ -----------------$\left(1\right)$

and $u=\sqrt{\frac{6}{7}}\times v$ ------------------$\left(2\right)$

putting both equation in $H$ and $h$ and solving after we get$,$

$\frac{2}{1}=\frac{6}{7}{\sin }^2\theta$

$\sin \theta =\sqrt{\frac{3}{7}}$

$\theta ={40}^0$

Hence,

option $\left(A\right)$ is correct answer.