Projectile velocity?

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 4.5 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.
HB/HA

When solving this problem I started out by working with the Ha part first and found the variables. X-R Vx-V y=Ha Voy a=-9.80m/s^2 t=?

Hb X=4.5R Vx=V Voy=0 a=-9.80 y=? t=?

Then i solved for time for Ha and for Hb
Time Ha=R/V
Time Hb=4.5R/V

then I used the kinematic equation and plugged in time for Ha and Hb

I ended up with 1/2*(9.80m/s^2)(R/V^2)
& Hb=-4.9m/s^2 (4.5R/V)^2


Im not sure what to do next...I know I need to find the missing variables but I'm not sure how to go about that! Please can you help me.

Answer 1

♦ if no gravity u would hit both times! Let initial speed of the bullet be v then
time t1=d1/v and t2=d2/v, where d2=4.5*d1,
♥ meanwhile HA=0.5g*t1^2 and HB=0.5g*t2^2, g=9.8m/s^2;
►HB/HA= t2^2 / t1^2 =(d2/d1)^2 =4.5^2;

Answer 2

Here's a hint.

Assuming the horizontal velocity of the bullet is constant, i.e. it does not slow down from its initial muzzle velocity, the only thing you need to calculate is the ratio of the distance something falls for one unit of time and the distance it will fall in 4.5 units of time. The horizontal velocity of the bullet doesn't matter.