Question

A horizontal rifle is lined up with and fired at bulls-eye target. The muzzle speed of the bullet is 680 m/s. The barrel is pointed directly at the center of the bulls-eye, but strikes the target 0.02 m below the center.

a) What is the bullet's inital velocity in the y-direction?

b) How long does the bullet take to reach the target ?

c) What is the horizontal distance between the end of the rifle and the bulls-eye?

d) What is the vertical y-velocity of the bullet when it hits the target ?

Answer 1

a) Since the rifle is horizontal, the initial velocity of bullet in y-direction is 0 m/s

b) the bullet travels 0.02 m in y-direction

dist = 0.02m

v_i = 0

a = acceleration due to gravity = 9.8m/s²

=> 0.02 = 9.8 * t²/2 => t² = 0.04/9.8 => t = 0.063s

the bullet takes 0.063s to reach the target

c)

here v_i = 680m/s

t = 0.063s

a = 0 as there is no acceleration in horizontal direction

so horizontal distance = 680 * 0.063 = 42.84m

d) v_f = v_i + at

v_i = 0, as initial velocity in y-direction is 0 m/s

a = acceleration due to gravity = 9.8 m/s²

t = 0.063s

v_f = 9.8 * 0.063 = 0.6174 m/s