Question

A car accelerating at the rate of $2m/s^2$ from rest is carrying a man at the rear end who has a gun in his hand. The car is always moving along + ve x-axis. At t = 4s, the man fires from the gun and the bullet hits a bird at t = 8s. The bird has a position vector:$40\hat{i}+80\hat{j}+\hat{k}$ at the time of firing the bullet. If velocity of project of bullet is $v=v_{x}i+v_{y}j+v_{z}k$. Take y axis in the horizontal plane. ___

Answer 1

Velocity of the car at the time of firing $=\frac{2}{4}=8m/s\hat{i}$. Now let us assume that velocity of projection of the bullet with respect to the car $v_x\hat{i}+v_y\hat{j}+v_z\hat{k}\dots$
Therefore, velocity of the bullet with respect to ground $=(v_x+8)\hat{i}+v_y\hat{j}+v_z\hat{k}$.
Now $(v_x+8)4=40\Rightarrow v_x=2m/s$;
$\left(v_y\right)4=80;v_y=20m/s;v_z\left(4\right)=120\Rightarrow v_z=30m/s)$
Hence, $\overrightarrow{v_s}=2\hat{i}+20\hat{j}+30\hat{k}$