Question

An archer shoots an arrow from a height $4.2\, m$ above the ground with a speed $40 \,m / s$ 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, $\left(\right.$ Take $\left.g=10\, m / s ^{2}\right)$

• 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=40 \,m / 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,
$\Rightarrow h=u t+\frac{1}{2} g t^{2}$
here, $h=4.2 m ,\, u=v \sin 30^{\circ}$
$4.2=v \sin 30^{\circ} t+\frac{1}{2} g t^{2}$
$4.2=40 \times \frac{1}{2} t+\frac{1}{2} \times 10 t^{2}$
$4.2=20\, t+5 t^{2}$
$50 t^{2}+200\, t-42=0$
$25 t^{2}+100 \,t-21=0$
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, $R=R_{1}+R_{2}$
$=80 \sqrt{3}+4 \sqrt{3}=84 \sqrt{3}\, m$