Question

A car accelerating at the rate of 2 $m/s^2$ from rest from origin is carrying a man at the rear end who has a gun in his hand. The car is always moving along positive x-axis. At t=4 s, the man fires from the gun and the bullet hits a bird at t=8 s. The bird has a position vector $40\hat{i}+80\hat{j}+40\hat{k}.$ Find velocity of projection of the bullet. Take the y-axis in the horizontal plane. $(g=10m/s^2)$

Answer 1

Given : Acceleration of car $a=2m/s^2$

Initial velocity of car $u=0$

$\therefore$ Distance moved by car in 4 s $\overrightarrow{S_c}=0+\frac{1}{2}\times 2\times 4^2=16\hat{i}$

Position vector of bird $\overrightarrow{r_b}=40\hat{i}+80\hat{j}+40\hat{k}$

$\therefore$ Position vector of bird w.r.t car $\overrightarrow{r_{bc}}=\overrightarrow{r_b}-\overrightarrow{S_c}=24\hat{i}+80\hat{j}+40\hat{k}$

Let the projection velocity of the bullet in car frame be $V$ making an angle $\theta$ with x axis in x-z plane.

$⟹$ $u_x=Vcos\theta$ and $u_z=Vsin\theta$

Time taken by bullet to hit the bird $t=8-4=4$ s

Car frame :

x direction : $S_x=u_{x}t$ where $S_x=24$

$\therefore$ $24=Vcos\theta \times 4$ $⟹Vcos\theta =6$ .................(1)

z direction : $S_z=u_{z}t+\frac{1}{2}{a}_z{t}^2$ where $S_z=40$ and $a_z=-g=-10m/s^2$

$\therefore$ $40=Vsin\theta \times 4-\frac{10\times 4^2}{2}$ $⟹Vsin\theta =30$ .................(2)

Squaring and adding (1) & (2) $V^2={30}^2+6^2=936$

$⟹$ $V=30.6m/s$ (in car frame)

Also from (2) /(1) we get $tan\theta =5$ $⟹\theta =ta{n}^{-1}5={78.7}^o$