Question

The maximam range of a bullet fired from a toy pistol mounted on a car at rest is R=40m. What will be the acute angle of inclination of the pistol for maximum range when car is moving in the direction of firing with uniform velocity v=20m/s, on the horizontal surface?

Answer 1

Maximum range of projectile is given by formula

$R = \frac{v^2}{g}$

now as per the formula

$40 = \frac{v^2}{10}$

$v = 20 m/s$

with respect to ground the velocity must make an angle of 45 degree

so its velocity components along X and Y direction must be

$v_x = 10\sqrt2$

$v_y = 10\sqrt2$

velocity with respect to car is given as

$v_x' = 10\sqrt2 - 20 = -5.86 m/s$

now in order to find the inclination

$\theta = tan^{-1}\frac{v_y}{v_x'}$

$\theta = tan^{-1}\frac{10\sqrt2}{5.86}$

$\theta = 67.5 degree$

so angle of inclination must be 67.5 degree