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 horizontal distance travelled before it hits the ground.
(Take $g=10m/s^2$)

• A. $80\text{m}$
• B. $48\text{m}$
• C. $64\text{m}$
• D. $32\text{m}$

Answer 1

The correct option is C $64\text{m}$

Using second equation of motion,
$s_y=u_{y}t-1/2g{t}^2$
$\Rightarrow -32=20\sin {37}^{\circ }t-5t^2$
$\Rightarrow -32=12t-5t^2$
$\Rightarrow 5t^2-20t+8t-32=0$
$\Rightarrow (5t+8)(t-4)=0$
$\therefore \text{Time of flight}t=4\text{s}$ (taking positive value of time)
Horizontal distance travelled depends only on the horizontal component of velocity and the horizontal component of velocity doesn’t change during the motion.
Hence, horizontal distance travelled by the body before it hits the ground $=u_x\times t$
$=20\cos {37}^{\circ }\times 4=20\times \frac{4}{5}\times 4=64\text{m}$