# Homework Help: Relative Velocity and Projectile Motion

1. Sep 19, 2008

### Kandycat

1. The problem statement, all variables and given/known data
A low-flying helicopter flying a constant 215 km/h horizontally wants to drop secret documents to his contact's open car which is traveling 155 km/h on a level highway 78.0 m below. At what angle (to the horizontal) should the car be in his sights when the packet is released?

2. Relevant equations
Kinematic Equations for Constant Acceleration in Two Dimensions
X-component (horizontal):
vx = vx0 + axt
x = x0 + vx0t + .5axt2
vx2 = vx02 + 2ax(x-x0)

Y-Component (vertical):
vy = vy0 + ayt
y = y0 + vy0t + .5ayt2
vy2 = vy02 + 2ay(y-y0)

3. The attempt at a solution
Knowns:
Vhelicopter = 59.7 m/s
Vcar = 43.1 m/s
y = -78 m
y0 = 0
x0 = 0
g = -9.8 m/s2

vy2 = ?
vy2 = vy02 + 2ay(y-y0)
vy2 = 0 + 2 (-9.8 m/s2)(-78 m)
vy = 39.1 m/s

vx = Vhelicopter + Vcar = 59.7 m/s + 43.1 m/s = 102.8 m/s ? <-- I'm not sure I did this right. Probably didn't.

tan Θ = vy/vx = 39.1/102.8
Θ = 20 degrees

But the back of the book says that the answer is 49.6 degrees. I'm probably wrong and somewhere in my work I screwed up somewhere. My calculator is in degree mode.

2. Sep 19, 2008

### Kurdt

Staff Emeritus
Ok try finding the x distance the parcel will travel and take the inverse tan of the ratio of the two sides (i.e. x - distance and y - distance).

3. Sep 19, 2008

### Kandycat

All right. I'm not sure if this is correct or not, but I found time.

y = y0 + vy0t + .5ayt2
-78 m = 0 + 0 + .5 (9.8 m/s2)t2
t = 3.99 s

x = x0 + vx0t + .5axt2
x = 0 + (59.7 m/s - 43.1 m/s)(3.99 s) + 0
x = 66.2 m

tan Θ = y/x = 78 m/66.2 m
Θ = 49.6 degrees

I'm still a bit confused about relative velocity though. Can someone explain how relative velocity works?

4. Sep 19, 2008

### Kurdt

Staff Emeritus
5. Sep 19, 2008

### Kandycat

Thank you so much for your help!