Question

The range of a rifle bullet is $1000m$, when $\theta$ is the angle of projection. If the bullet is fired with the same angle of projection from a car travelling horizontally at 36km/hr towards the target, the increase in the range will be

$(given\sqrt{\frac{5}{9.8}}=0.712)$

• A. $173\sqrt{tan\theta }$
• B. $183\sqrt{tan\theta }$
• C. $143\sqrt{tan\theta }$
• D. $113\sqrt{tan\theta }$

Answer 1

Answer: (C)

$143\sqrt{tan\theta }$

Velocity of car $=36km/hr$ $=10m/s$
initial range $R=\frac{2u^{2}sin\theta cos\theta }{g}=\frac{2}{g}PQ$
where $P=ucos\theta$, $Q=usin\theta$
Final range $R^{{}^{\prime }}=\frac{2}{g}(P+10)Q$
Since vertical component remains the same
Increase in range, $R^{{}^{\prime }}-R=\frac{20Q}{g}=\frac{20usin\theta }{g}$
$R=\frac{2u^{2}sin\theta cos\theta }{g}=1000$ (given)
$u^2=\frac{1000\times g}{2sin\theta cos\theta }$
Solving (1) and (2)
$\Rightarrow R^{{}^{\prime }}-R=\frac{20\sqrt{1000}}{\sqrt{g}}\times \frac{\sqrt{tan\theta }}{\sqrt{2}}=143\sqrt{tan\theta }$