Question

An archer shoots an arrow from a height $4.2m$ above the ground with a speed $40m/s$ and at an angle $30^{\circ }$ as shown in the figure. Determine the horizontal distance $R$ covered by the arrow, when it hits the ground, $($ Take $g=10m/s^2)$

• A. $\frac{185}{\sqrt{3}}m$
• B. $84\sqrt{3}m$
• C. $68\sqrt{3}m$
• D. $\frac{95}{\sqrt{3}}m$

Answer 1

Answer: (B)

Given, speed of arrow,
$v=40m/s$ and $\theta =30^{\circ }$

Horizontal range $R_1$ covered by the arrow is given by
$R_1=\frac{v^2\sin 2\theta }{g}$
$=\frac{40^2\sin (2\times 30)}{10}=160\sin 60^{\circ }$
$=160\frac{\sqrt{3}}{2}=80\sqrt{3}m$
If $t$ be the time taken by the arrow reaching from $B$ to $C$, then from the second equation of motion,
$\Rightarrow h=ut+\frac{1}{2}g{t}^2$
here, $h=4.2m,u=v\sin 30^{\circ }$
$4.2=v\sin 30^{\circ }t+\frac{1}{2}g{t}^2$
$4.2=40\times \frac{1}{2}t+\frac{1}{2}\times 10t^2$
$4.2=20t+5t^2$
$50t^2+200t-42=0$
$25t^2+100t-21=0$
Solving the quadratic equation, $t=\frac{1}{5}s$
Distance travelled in horizontal direction is given by
$R_2=v\cos 30^{\circ }\times t$
$=40\frac{\sqrt{3}}{2}\times \frac{1}{5}=4\sqrt{3}m$
Hence, the horizontal distance covered by the arrow, $R=R_1+R_2$
$=80\sqrt{3}+4\sqrt{3}=84\sqrt{3}m$