Physics, 11.07.2019 15:00 maddy3lizabeth

Arifle 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.

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The ratio of the distances $h_{b}/h_{a}$ of a bullet that is shot twice to a target is 12.25, calculated knowing that the second time, the rifle aimed from 3.5 times the distance from the target.

The vertical distance for the semi parabolic movement can be calculated with the following equation:

$h_{f} = h_{i} + v_{i_{y}}t - \frac{1}{2}gt^{2}$

Where:

$h_{f}$ : is the final vertical distance

$h_{i}$ : is the initial vertical distance

$v_{i_{y}}$ : is the initial velocity of the bullet in the y-direction

t: is the time

g: is the acceleration due to gravity = 9.81 m/s²

When the rifle shoots the target the first time, we have:

$0 = h_{a} + 0*t - \frac{1}{2}gt_{1}^{2}$

$h_{a} = \frac{1}{2}gt_{1}^{2}$ (1)

The initial velocity in the y-direction is zero (0) because the bullet is initially moving horizontally (it has an initial velocity in the x-direction).

Now, when the bullet is shooted a second time, we have:

$h_{b} = \frac{1}{2}gt_{2}^{2}$ (2)

We can find the ratio of $h_{b}/h_{a}$ by dividing equation (2) by equation (1):

$\frac{h_{b}}{h_{a}} = \frac{\frac{1}{2}gt_{2}^{2}}{\frac{1}{2}gt_{1}^{2}} = (\frac{t_{2}}{t_{1}})^{2}$ (3)

The times t₁ and t₂ can be calculated knowing that, the second time, the rifle aimed from 3.5 times the distance from the target. We can use the next equation:

$d = v_{i_{x}}t$

Where:

d: is the distance

$v_{i_{x}}$ : is the initial velocity of the bullet in the x-direction

For the first time, we have:

$d_{a} = v_{i_{x}}t_{1}$ (4)

And for the second:

$d_{b} = v_{i_{x}}t_{2}$ (5)

By dividing equation (5) by equation (4) we can find the ratio t₂/t₁:

$\frac{d_{b}}{d_{a}} = \frac{v_{i_{x}}t_{2}}{v_{i_{x}}t_{1}} = \frac{t_{2}}{t_{1}}$

Since $d_{b} = 3.5d_{a}$ we have:

$\frac{t_{2}}{t_{1}} = \frac{3.5d_{a}}{d_{a}} = 3.5$ (6)

Finally, we can find the ratio $h_{b}/h_{a}$ after entering equation (6) into eq (3):

$\frac{h_{b}}{h_{a}} = (\frac{t_{2}}{t_{1}})^{2} = (3.5)^{2} = 12.25$

Therefore, the ratio of $h_{b}/h_{a}$ is 12.25.

You can find more about velocity and displacement in a semi parabolic motion here:

linklink

I hope it helps you!

Answer from: ueuwuwj

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.