Question

The horizontal distance travelled by a bullet fired at ${45}^{\circ }$ is $360m$. If it is fired from a lorry moving in the direction of bullet with the uniform velocity $18km{h}^{-1}$ and with the same elevation, what is the horizontal distance travelled by the bullet? (Take $g=10m{s}^{-2}).$

• A. $402.42km$
• B. $201km$
• C. $402.42m$
• D. $201m$

Answer 1

Answer: (C)

$402.42m$

Range = $\frac{u^{2}sin2\theta }{g}=\frac{u^2\times sin{90}^0}{g}=\frac{u^2}{g}=360$
So, $u=60m/s$
With lorry,
Time period = $\frac{2u\times sin\theta }{g}=(2\times 60\times (\frac{1}{\sqrt{2}})\times \frac{1}{10})=\frac{12}{\sqrt{2}}s$
Range = $v_{x}t=(60cos\theta +5)t$
= $(\frac{60}{\sqrt{2}}+5)\frac{12}{\sqrt{2}}$
Therefore, the horizontal distance travelled by the bullet $=402.42m$