posted by Sarah .
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
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
where a=-g, and u the muzzle velocity.
The formula for S (distance) applies independently to horizontal and vertical directions.
For part (a),
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
t=50 yards * 3 ft/yd / 750 ft/s
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.
how did you .644 to not be a negative ???
Correct, it should be negative, since the bullet drops downwards.
well in the book it says its not negative either
You have the correct mathematical answer of -0.644.
But the question asks
"How much does the bullet drop in flight ..."
so the direction is implied, so the negative is not required.
You only have to translate the mathematical answer to a human answer.