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 $v_x=20\cos {37}^{\circ }=16\text{m/s}$
Final vertical velocity is given by
$v_y^2=u_y^2+2gs{v}_y^2=(20\sin {37}^{\circ }{)}^2+2\times (-10)\times (-32)v_y^2=144+640=784$
$\Rightarrow v_y=-28\text{m/s}$ (negative sign indicates downward direction)
Therefore, final velocity of the body
$v=\sqrt{v_x^2+v_y^2}=\sqrt{{16}^2+(-28{)}^2}=\sqrt{1040}=32.25m/s$