Relative Velocity and Projectile Motion

[Kandycat]

Homework Statement

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?

Homework Equations

Kinematic Equations for Constant Acceleration in Two Dimensions
X-component (horizontal):
v_x = v_x0 + a_xt
x = x0 + vx0t + .5axt2
v_x² = v_x0² + 2a_x(x-x₀)

Y-Component (vertical):
v_y = v_y0 + a_yt
y = y₀ + v_y0t + .5a_yt²
v_y² = v_y0² + 2a_y(y-y₀)

The Attempt at a Solution

Knowns:
V_helicopter = 59.7 m/s
V_car = 43.1 m/s
y = -78 m
y₀ = 0
x₀ = 0
g = -9.8 m/s²

v_y² = ?
v_y² = v_y0² + 2a_y(y-y₀)
v_y² = 0 + 2 (-9.8 m/s²)(-78 m)
v_y = 39.1 m/s

v_x = V_helicopter + V_car = 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 = v_y/v_x = 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.

 

[Kurdt]

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).

 

[Kandycat]

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

y = y₀ + v_y0t + .5a_yt²
-78 m = 0 + 0 + .5 (9.8 m/s²)t²
t = 3.99 s

Then after I found time. I took your advice.

x = x₀ + v_x0t + .5a_xt²
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?

 

[Kurdt]

Perhaps this page will help things. If you have any more specific questions afterwards feel free to ask.

http://hyperphysics.phy-astr.gsu.edu/Hbase/relmot.html

 

[Kandycat]

Thank you so much for your help!