Posted by Sarah on Sunday, June 7, 2009 at 9:53pm.
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 t^{2}/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 t^{2}/2
= 0 + (-32.2)0.2^{2}/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.
Thank you.
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.