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Question

A shotgun is fired horizontally at a target in a firing range. The bullet leaves the gun at 598 m/s. The gun is pointed directly at the center of the target, but the bullet strikes the target 0.032 m below the center. How far is the shotgun from the target? Include a sketch that shows the trajectory of the bullet as a dotted line, and labeled vectors for ∆x, ∆y, and v_i.

Answer 1

Let

v = 598 m/s be the bullet's speed

h = 0.032 m be the distance between the target's center and the impact site,

g = 9.8 m/s² be gravitational acceleration.

The bullet followed a parabola resulting from a combination of two movements:

- A horizontal trajectory due to inertia, which is what the bullet would follow in the absence of gravity.

- A vertical fall due to gravity, which is what the bullet would follow in the absence of any initial horizontal velocity.

You can view the trajectory in an X-Y coordinate space, where

the X coordinate is horizontal at the level of the gun, and

the Y coordinate is vertical flush against the muzzle of the gun.

Then [x(t), y(t)] is the position of the bullet after time t.

(We start measuring time t the instant the bullet leaves the gun.)

x(t) = vt

y(t) = -gt²/2

The fall continues until the bullet hits the target at some time t₁,

which we can calculate because at time t₁, y(t₁) = -h. That is,

- gt₁²/2 = -h

t₁ = sqrt(2h/g) = sqrt(0.064 m/(9.8 m/s²)) = 0.08 s

Having obtained t₁, we can calculate how far the bullet flew horizontally by:

x(t₁) = vt₁ = 598 m/s x 0.08 s = 48.3 m