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## Homework Statement

Bruce Willis is driving his car up a 30° ramp in an attempt to crash into a helicopter. The ramp is 2 meters high, the helicopter is 10 meters away from the edge of the ramp and 6 meters from the ground.

Find the speed of his car when it leaves the ramp and time he is flying through the air.

## Homework Equations

[tex]x = x_0 + v_0 cos(30°) t[/tex]

[tex]y = y_0 + v_0 sin(30°) t - \frac{1}{2} gt^2[/tex]

## The Attempt at a Solution

To put it simply, I have two variables and two equations. Simple enough.

[tex]t = \frac{10m}{v_0 cos(30°)}[/tex]

Putting t into the latter equation I get.

[tex]8m = 2m + v_0 sin(30°)\ \frac{10m}{v_0 cos(30°)}\ -\ \frac{1}{2}\ 9.8\frac{m}{s^2}\ (\frac{10m}{v_0 cos(30°)}^2)[/tex]

[tex]v_0 = \sqrt{\frac{428.3 m^2}{3 s^2}} = 11.95m/s = 43 km/hour[/tex]

This is where I freeze up. I'm 98% sure that what I'm doing is right. Yet 43 km/hour for a car to get a helicopter 4 meters above you and 10 meters away, that sounds dodgy.

Also, If I increase the height of the helicopter above the ground I get a negative under root. But that's probably because the helicopter is so far above that no matter how fast you're going, you would just go under it in a straight line.

Is this correct?

Thanks for your help.