Question

A projectile projected from the ground has its direction of motion making an angle $\frac{\pi }{4}$ with the horizontal at a height $40m$. If initial velocity of projection is $50m/s$, the angle of projection is (take $g=10m{s}^{-2}$)

• A. $\frac{1}{2}{\cos }^{-1}(-\frac{8}{25})$
• B. $\frac{1}{2}{\cos }^{-1}\left(\frac{8}{25}\right)$
• C. $\frac{1}{2}{\cos }^{-1}(-\frac{4}{5})$
• D. $\frac{1}{2}{\cos }^{-1}(-\frac{1}{4})$

Answer 1

The correct option is A

$\frac{1}{2}{\cos }^{-1}(-\frac{8}{25})$

Let projection angle be $\theta$ and velocity at $40m$ height be '$v$'

then, $50\cos \theta =v\cos {45}^o$

$\Rightarrow v=50\sqrt{2}\cos \theta$ ............. (1)

Now, $v_y^2-u_y^2=2a_y{s}_y$

$\Rightarrow (v\sin {45}^o{)}^2-(50\sin \theta {)}^2=2(-10)\left(40\right)$ .........(2)

From (1) and (2)

$\theta =\frac{1}{2}{\cos }^{-1}(-\frac{8}{25})$