# Physics

A markman fires a .22 caliber rifle horizontally at a target the bullet has a muzzle velocity with magnitude 750 ft/s. How much does the bullet drop in flight if the target is (a) 50.0 yd away and (b) 150.0 yd away?

for this question i was given the equation
S= ut + at^2/2

I don't know what numbers I should put where all i know is that a = -9.8 and u = 750 but the answer for the question is .64 how do they get this

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1. Sorry Sarah if the previous explanation was not clear.

Since he shoots horizontally, the drop starts the moment the bullet leaves the rifle, at the rate of g (acceleration due to gravity).
For (a) and (b), calculate the time t the bullet is in the air using the horizontal speed, then calculate drop using
S=ut+at^2/2
where a=-g, and u the muzzle velocity.

The formula for S (distance) applies independently to horizontal and vertical directions.
u=initial velocity
a=acceleration
t=time

For part (a),
Horizontal direction:
time to reach 50 yards is the distance divided by the muzzle velocity. There is no acceleration (a=0) and air resistance is neglected.
S=ut + a t2/2
50*3 = 750 * t + 0
Thus
t=50 yards * 3 ft/yd / 750 ft/s
=0.2 s.

Vertical direction:
u=initial vertical velocity = 0 (bullet was shot horizontally)
a=-g = -32.2 ft/s/s
S = 0*t + a t2/2
= 0 + (-32.2)0.22/2 ft.
= 0 + (-32.2)* 0.04 /2 ft.
= 0.644 ft.

I am sure you can do part (b) along the same lines. Post your answer if you want it checked.

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2. how did you .644 to not be a negative ???

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3. Correct, it should be negative, since the bullet drops downwards.
Thank you.

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4. well in the book it says its not negative either

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5. You have the correct mathematical answer of -0.644.