How to Solve a Tough Concept Question: Rifle Aiming Experiment

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The discussion centers on solving a physics problem involving a rifle aimed at a target from two different distances. The first shot hits a point HA below the bull's eye, while the second shot, taken from 3.13 times the distance, hits HB below the center. Participants clarify that the time of flight for the second bullet is 3.13 times longer than for the first. By applying the equation for distance fallen due to gravity, the ratio HB/HA can be determined using the relationship of time squared. Ultimately, the solution hinges on understanding the proportionality of the time of flight to the vertical drop of the bullets.
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Tough Concept Question!

Homework Statement


A rifle is used to shoot twice at a target using identical cartridges. The first time, it 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 3.13 times the distance from the target. This time the bullet strikes the target at a distance of HB below the center. Find the ratio HB/HA.


Homework Equations


N/A


The Attempt at a Solution


this is a very tough question i have come upon, i tried maybe going 2 x 3.13 as the Hb and dividing it by 2 but it doesn't seem to work, anyone have any ideas??
 
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HA = 1/2gt^2

HB = 1/2gT^2

s = ut
3.13s = uT

T/t = 3.13

... and so on
 


^^^^^
could u perhaps clarify that?
 


mal4mac is demonstrating that the time the bullet from rifle 2 is traveling through the air is 3.13 times the amount of time the bullet from rifle 1 traveled through the air.
Now all you need to do is apply x = .5*g*t^2 to find the distance the bullet dropped iduring its flight.
All you need is a ratio, so:
x1 = .5*g*(t1^2)
x2 = .5*g*(t2^2)

or:

x2/x1 = t2^2 / t1^2
Since t2 is 3.13x greater than t1, plug this in and you have your ratio.
 
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