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 velocity with which it hits the ground.

(Take $g=10m/s^2$)

• A. $32.25\text{m/s}$
• B. $16\text{m/s}$
• C. $20\text{m/s}$
• D. $28\text{m/s}$

Answer 1

The correct option is A $32.25\text{m/s}$

Since the horizontal component of velocity does not change,
Final horizontal velocity = initial horizontal velocity
i.e $v_x=20\cos {37}^{\circ }=16\text{m/s}$

Initial vertical velocity = $u_y=20sin37=12m/s$

Final vertical velocity is given by
$v_y^2=u_y^2+2gs=(12{)}^2+2\times 10\times 32=28\text{m/s}$ (in downward direction)
Therefore,the magnitude of final velocity of the body
$v=\sqrt{v_x^2+v_y^2}=\sqrt{{16}^2+(28{)}^2}=\sqrt{1040}=32.25m/s$

Direction-

$\tan \alpha =\frac{28}{16}=\frac{7}{4}$

$\alpha ={\tan }^{-1}\frac{7}{4}$
where $\alpha$ is angle with the horizontal