How do i solve this physics projectile motion problem?

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 to the center of the bull's eye. The bullet strikes the target at a distance Ha below the center, however. The second time, the rifle is similarly aimed but twice the distance from the target. This time the bullet strikes at a distance of Hb below the center. Find the ratio of Ha/Ba

Answer 1

Let X be the distance to the target, and assume no air resistance so that muzzle velocity Vx is constant (and the same for both parts).

Case a)

Time to hit target:

X = Vx * Ta ==> Ta = X / Vx

Case b)

Time to hit target:

2 * X = Vx * Tb ==> Tb = 2 * X / Vx

These are the same times that the projectile has to fall under gravity:

H = 1/2 * g * T^2, so

Ha = 1/2 * g * [X / Vx]^2

Hb = 1/2 * g * [2 * X / Vx]^2

in the ratio Ha / Hb, g, X, and Vx cancel out leaving

Ha / Hb = 1/4

Good Luck!

Answer 2

Assuming we can neglect air resistance, as most problems do:

The bullet accelerates downward at 9.8 m/s^2, starting with an initial downward velocity Vi of 0.

In a period of time T, the bullet will reach a final downward velocity of: Vf = (9.8 * T)

The distance traveled in that time will be the average vertical speed multiplied by time: Ha = [(Vf + Vi)/2] * T.

Since Vi = 0, we can simplify: Ha = 1/2 * (Vf * T)

When you double the distance, you are also doubling the time before the bullet hits the target (if we ignore air resistance).  This gives the bullet twice as much time to fall.  But it will fall more than just twice as much because it is accelerating.  It will accelerate to: (2 * Vf).

So the 2nd equation is:

Hb = [ (2Vf + 0) / 2 ] * 2T = 2 (Vf * T)

Comparing the two equations is the same as comparing 1/2 and 2.  The ratio (Ha / Hb) = .5 / 2 = 1/4

So your answer is 1/4.