Trajectory of a cyclist leaving a ramp

[pb23me]

Homework Statement

The problem says: A motorcycle daredevil is attempting to jump across 10 buses. The takeoff ramp makes an angle theta = 18 degrees above the horizontal, and the landing ramp is identical to the take off ramp. The buses are parked side by side and each bus is 2.74m wide. The cyclist leaves the ramp with a speed of v = 33.0. does he succeed in jumping over the buses? I am not sure how to go about solving this prob. It seems as though I am supposed to assume Yf=0 but i don't think i should?



Now my variables:
Xi : 0
Xf:?
Ax : 0 m/s^2
Vxi : 33.0(cos 18. degrees)
Vxf : 33.0(cos 18. degrees)
t : ?

Yi:0
Yf:?
Ay : ?
Vyf : ?
Vyi : 33.0(sin 18 degrees)
t : ?

Homework Equations

The Attempt at a Solution

 

[Delphi51]

Yes, I would assume assume Yf=0.
The height of the buses is not given, so I would assume the ramps are sufficiently high that the bike clears the first and last buses okay and the gap between ramps is just 10*2.74 meters.

You should be able to get the time of flight easily from the horizontal part of the motion.

 

[pb23me]

why do you assume the bike hits the ramp at exactly the same y coordinate as when it took off? it seems to me you wouldn't really know that eg it hits much lower down the ramp.

 

[Liquidxlax]

find the y component velocity and see how long it takes to reduce vertical velocity to zero due to gravity. And you know that it takes the same amount of time to fall that same distance. Does horizontal velocity ever change? with knowing this you know how far it hits on the ground

 

[pb23me]

no horizontal doesnt. and so your saying find time for the motorcycle to reach max height then multiply that time by two??

 

[Liquidxlax]

no horizontal doesnt. and so your saying find time for the motorcycle to reach max height then multiply that time by two??

well you have to know why the vertical path has a total time of 2t.

but yes time is multiplied by 2

 

[Delphi51]

Sorry, I was thinking of a slightly different problem!
You are quite right, there is something missing from this question.
As it stands, the bike will cross the horizontal distance of 27.4 m in 0.873 seconds. In that time the vertical motion will be incomplete and the bike will be more than 6 meters above the launch point. If the ramps were moved back away from the buses to synchronize the horizontal and vertical motion, it could work. Question not clear?

 

[Liquidxlax]

Sorry, I was thinking of a slightly different problem!
You are quite right, there is something missing from this question.
As it stands, the bike will cross the horizontal distance of 27.4 m in 0.873 seconds. In that time the vertical motion will be incomplete and the bike will be more than 6 meters above the launch point. If the ramps were moved back away from the buses to synchronize the horizontal and vertical motion, it could work. Question not clear?

technically it does not state he has to hit the landing ramp, it asks if he will jump all the buses

 

[Delphi51]

Good point! So the answer is very simple after all.
Kind of a wreck of a question, though.