Bruce Willis vs. helicopter bad guy

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Homework Help Overview

The problem involves a scenario where Bruce Willis is attempting to drive a car off a ramp to reach a helicopter positioned at a certain height and distance. The subject area includes kinematics and projectile motion, focusing on the calculations of speed and time of flight.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the equations of motion and the variables involved. There are questions about the interpretation of time and distance, particularly regarding the height of the helicopter and the implications of the calculated speed.

Discussion Status

Participants are actively engaging with the problem, offering different interpretations and calculations. Some suggest alternative approaches to finding the time of flight and speed, while others express uncertainty about the correctness of their results. There is no explicit consensus, but productive dialogue is occurring around the calculations.

Contextual Notes

There are indications of potential confusion regarding the problem's parameters, such as the height of the helicopter and the distance involved. Some participants note the need to clarify the question's requirements, particularly concerning the time of flight versus the time to reach the helicopter.

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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.
 
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I think i know why you're getting some issues. Is it possible that the question is asking what the time flying through the air is and instead, you're finding the time is takes to get to the helicopter??
 
ive done it and the answer i get is approx 19m/s, which looks to me like it could be right.

looking at your answer, i don't see where you get the 8m in there. But if you solve for v in the x equation, then sub that into the y equation to find t, then sub that t back into the x equation, i don't really see where you could go wrong.
 
and you are right about the negative root. unless you put some wings on the car and calculate some thrust:)
 
dacruick said:
I think i know why you're getting some issues. Is it possible that the question is asking what the time flying through the air is and instead, you're finding the time is takes to get to the helicopter??

The time to get to the helicopter is the one I need. Sorry about that. And those 8m are supposed to be 10m. My bad again.

19m/s is a lot better. I went the other way and solved the time from the x-equation and put it in the y-equation to solve V_0. Don't know what went wrong.

Cheers. :)
 
either way you can find the time with that 19 metres. with that 19m you have initial velocity and an initial angle, you can definitely find the time in the air from that.
 

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