# Projectile Motion ~ AP Physics (A rifle is fired at an angle of )

Silverbolt
A rifle is fired at an angle of 25° below the horizontal with an initial velocity of 400 m/s from the top of a cliff 60.0m high. How far from the base of the cliff does the bullet land?

This is what I did, but i'm not sure if its right:

y-component→ v=400sin(25°) ∴v=169.05
x-component→ v=400cos(25°)∴v=362.52
I found the time by doing: t= 60/169.05∴t= .355s
From here I multiplied the time with the x-component to find the distance(X) : X=.355(362.52)∴ X=128.69m

Is this right???

SHISHKABOB
don't forget about the constant acceleration due to gravity in the y-direction

Silverbolt
Could you explain a bit more, @SISHKABOB

SHISHKABOB
what I mean is that the time that you determined is the time that it would take an object to travel 60m while going a constant 169.05 m/s

but in the y direction, there is a constant acceleration due to gravity, so the projectile will be speeding up as it falls. It will start at 169.05 m/s, but will accelerate until it hits the ground.

You need to take this into account.

Silverbolt
@SISHKABOB, So you mean I should use this equation: Y(distance)=vt+.5at²
so that would be: 60=(169.05)t+.5(-9.8)t² therefore t=.359s

SHISHKABOB
@SISHKABOB, So you mean I should use this equation: Y(distance)=vt+.5at²
so that would be: 60=(169.05)t+.5(-9.8)t² therefore t=.359s

yep. It's a very small difference but that's only because the problem is talking about a bullet shot from a rifle. In slower situations it makes a much bigger difference.

try setting the initial velocity to like, 1 m/s, and then find the difference in time if you did it first ignoring gravity, and then considering gravity

Silverbolt
I see what you mean. Thank you!

willem2
@SISHKABOB, So you mean I should use this equation: Y(distance)=vt+.5at²
so that would be: 60=(169.05)t+.5(-9.8)t² therefore t=.359s

the bullet is shot downwards, and the acceleration from gravity is also downwards, so these should have the same sign. If the top of the cliff is at 0, the bottom should be at -60, and you should get -60 = (-169.05)t + (0.5) (-9.8)t² => 60=(169.05)t+.5(9.8)t

and the bullet should hit the ground a little faster than in the unaccelerated case, instead of slower.

SHISHKABOB
the bullet is shot downwards, and the acceleration from gravity is also downwards, so these should have the same sign. If the top of the cliff is at 0, the bottom should be at -60, and you should get -60 = (-169.05)t + (0.5) (-9.8)t² => 60=(169.05)t+.5(9.8)t

and the bullet should hit the ground a little faster than in the unaccelerated case, instead of slower.

oops! you're absolutely right, I wasn't thinking it through