Question

A rifle is used to shoot twice at a target, using identical cartridges. the first time, the rifle is aimed parallel to the ground and directly at the center of the bull's eye. the bullet strikes the target at a distance of ha below the center, however. the second time, the rifle is similarly aimed, but from 3.5 times the distance from the target. this time the bullet strikes the target a distance of hb below the center. find the ratio of hb/ha.

Answer 1

$h_b/ h_a = 49/4 = 12.25$

Assuming negligible air resistance, the shell travels at a constant horizontal velocity equal to what it had immediately after leaving the barrel. The shell is in the air until hitting the target; the length of this period of time is therefore directly proportional to the distance between the barrel and the target.

The shell hits the target below where it was aimed at for it experiences gravitational pull during its flight. With the barrel pointed "parallel to the ground" the shell shall experience no initial vertical velocity, meaning that it its undergoing a freefall on the vertical direction. $h_a$ and $h_b$ are therefore the respective vertical displacements of the two trials.

The formula below relates the vertical displacement to the gravitational acceleration $g$ and the time duration of the freefall $t$:

$\Delta h = 1/2 \cdot g \cdot t^{2}$

Let $t_a$ and $t_b$ be the duration of the shell's flight on the first and second trials, respectively; Thus

• $h_a = 1/2\cdot g \cdot {t_a}^{2}$
• $h_b = 1/2\cdot g \cdot {t_b}^{2}$

$h_b / h_a = (t_b / t_a)^2 = (3.5)^2 = 49/4 = 12.25$