Physics Help, 2 dimensional problems!?

Question

Problem 6.65
You have been hired to assist with stunts for a new action movie. In one scene, the writers want to drop a package from a small plane into a moving convertible sports car. The car will drive along a horizontal road at 30.0 m/s. The plane will approach the car from behind at an altitude of 60.0 m and a speed of 90.0 m/s relative to the ground. The copilot will view the car through a sighting tube that measures the angle below horizontal.

At what angle should the package be released? degrees below the horizontal

The plane and car lie along the same vertical axis.

Answer 1

16°

Projectile motion equations dictate that the horizontal velocity will remain constant (neglect air resistance). The distance travelled by the package will be:

x=vt, where v is the relative velocity of the car and the plane = 60m/s.

x=60t

The time to fall the distance to the ground is given by:
y=-1/2gt^2
-60=-1/2(9.81)t^2
t^2=120/9.81
t=3.5s (this is how long the package will take to hit the ground)

sub back in to get

x=60t=60(3.5)=209.8m (this is how far the package will travel horizontally)

to find the angle, simple triangle trig gives you a 60m vertical drop over a distance of 209.8m:

tan(A)=60/209.8
A=16° below the horizontal.

m

Answer 2

Ok, so basically, since the plane and the package will be moving at the same speed laterally, we want the package to hit the ground when the plane is over the car. Note that I use 10 for gravity... replace with 9.81... if you want.

The package will take A seconds to fall:
s = 1/2 g t^2
60m = 1/2*10*A^2
12 = A^2, A = sqrt(12) seconds.

So, we want the distance behind the car that the plane will be at sqrt(12) seconds before it overtakes it. Since the overtake speed is 60m/s, this is simpley 60*sqrt(12) (simplify sqrt(12) to 2sqrt(3)) or 120sqrt(3) meters.

For the down angle, the opposite side is the height of the plane (60m) and the adjacent side is the distance back (120sqrt(3)m), so the tangent of the down angle is 60/120sqrt(3) = 1/(2sqrt(3)).

So the arctangent of that should give the angle =~ 16 degrees.

There ya go.

Answer 3

Is this assuming that the car and the plane have to lie long the same vertical axis when the package is dropped?

Answer 4

The answer to the question is 30 degrees off vertical, you do the
conversion for horizontal.

On the other hand, speaking as a director, i would never put my
stunt people in that much danger. I would do the scene
completely in CG. It would be cheaper than the hospital bill or
the funeral.

Thank you very much, while you're up!

Answer 5

You need to state your assumpions here. Are you operating in a vacuum? What is the force from the wind. 60 miles is in space, so how is the airplane supposed to get up there? Where on the planet is this happening. You have chosen some conditions that would make it necessary to consider things like wind, variable atmospheric density, the spin and momentum of the Earth, force of gravity, things like that.

Answer 6

Hint: the answer is the same as if the plane were moving at 60.0 m/s and the car were stationary.

Answer 7

1.113 degrees below the horizontal.


Doug