# Sniper Projectile Motion

1. Jan 28, 2013

### Inferior Mind

A sniper fires a bullet at 120 m/s at 30° above the horizontal from the roof top of a 35 m high parking garage. If the bullet strikes the level ground beside the parking garage:

How long was the bullet in the air?

How far from the base of the parking garage did the bullet land?

At what angle did the bullet land?

g= -9.8 m/s^2
Vi= 120 m/s @ 30 degree incline
h=35 m
Vx= 120Cos30 = 140 m/s
Vy= 120Sin30 = 60 m/s

2. Jan 28, 2013

### SignaturePF

Use the following equation to find time:
Δy = v_oyt + 1/2at^2,
where Δy = - 35, v_0y = 60, a = -9.8m/s^2

3. Jan 28, 2013

### Inferior Mind

First off, doesn't a Muzzle Velocity seem a bit low at 120 m/s? -_-

Alright, so it looks like I end up with a Quadratic and solve for both values.

0 = -4.9t^2 + 60t + 35

t = -.55s and 12.8s

But there can't be negative time, amiright ?

And I am unsure about the Total change in the horizontal distance.

Would ~ Dx = Vx(t) ? [x represents the change in the horizontal or x-axis]

= (104 m/s)12.8 s
= 1,331.2 m

4. Jan 28, 2013

### SignaturePF

Ok good, I'm assuming you did the quadratic right.
If t is 12.8 s.
Simply use the same equation again:
Δx = v_oxt + 1/2at^2,
Here a = 0, so the equation becomes:
delta x = v_oxt
You have both of those.
It seems like you've done it right.
Good job.

5. Jan 28, 2013

### HallsofIvy

Staff Emeritus
It's uses a spring to fire the bullet!

The point was to separate horizontal (x) and vertical (y) components.

With straight line constant acceleration a, velocity is $at+ v_0$ and distance traveled is $(1/2)at^2+ v_0t+ s_0[itex] In this problem, vertically, the acceleration is -g= -9.82 m/s^2, approximately, so that [itex]y= -4.91t^2+ 60t+ 35$ where I have chosen the "0" height to be on the ground, at the bottom of the building. The bullet hits the ground, then, when $-4.91t^2+ 60t+ 35= 0$. Yes, that will have two roots, one negative and one positive. Yes, the positive root is the one you want.

Horizontally, the acceleration is 0 so the equation is $x= 140t$ where I have chosen x= 0 as the starting point, horizontally. To find how far the bullet goes horizontally, set t equal to the value you got before for the time the bullet is in the air.

6. Jan 28, 2013

### Inferior Mind

Alright, so assuming everything so far is correct, my only issue has been the angle at which the bullet hits the ground.

If Tan θ = O / A then could I use the components for Sin and Cos ?

My text says that I should change the Velocity of the Vertical Component by using:

Vr = Vy + aΔt
= 60 m/s + (-9.8 m/s)12.8s
= -65.5 m/s

Since this becomes the Hypotenuse of the triangle at the moment of impact it becomes positive.

... so Tan^-1(104/65.5) = 57.79° from the Vertical.

Can someone double check if this is an acceptable answer?