Question

A body is projected from the top of a tower of height $32\text{m}$ with a velocity of $20\text{m/s}$ at an angle of ${37}^{\circ }$ above the horizontal. Find the angle below the horizontal with which it hits the ground.
(Take $g=10m/s^2$)

• A. ${\tan }^{-1}\left(\frac{-3}{4}\right)$
• B. ${\tan }^{-1}\left(\frac{3}{4}\right)$
• C. ${\tan }^{-1}\left(\frac{7}{4}\right)$
• D. ${\tan }^{-1}\left(\frac{-7}{4}\right)$

Answer 1

The correct option is D ${\tan }^{-1}\left(\frac{-7}{4}\right)$

Horizontal velocity doesn't change.
$\Rightarrow v_x=u_x=u\cos {37}^{\circ }=16\text{m/s}$
Initial vertical velocity $u_y=20\sin {37}^{\circ }=12\text{m/s}$
Displacement at the time of hitting ground $s_y=-32\text{m}$
Using third equation of motion,
$v_y^2-u_y^2=2g{s}_y$
$v_y^2-{12}^2=2\times -10\times -32$
$\Rightarrow v_y=-28\text{m/s}$ (velocity is downward)
If $\theta$ is the angle at which the body hits the ground, then
$\tan \theta =\frac{v_y}{v_x}=\frac{-28}{16}=-\frac{7}{4}$
$\therefore \theta ={\tan }^{-1}(-\frac{7}{4})$
is the angle w.r.t the horizontal at which the body hits the ground.