Projectile Motion test problem, involving unk. v0 and y displacement.

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SUMMARY

The discussion revolves around a projectile motion problem where a stuntman must jump across a 40m wide river from a ramp inclined at 53 degrees and 100m high. The challenge is to determine the minimum initial velocity (v0) required to reach the opposite bank, which is 15m lower than the ramp. Key equations include the vertical displacement equation (Δy = v0yt - 0.5gt²) and the range equation (R = (v²sin2Θ)/g). Participants suggest using a variable for initial speed and solving the problem as a system of equations to find v0.

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Homework Statement


A movie set requires a stunt man to jump across a 40m wide river on a motorcycle. The takeoff ramp on one bank of the river is inclined at 53(degrees) above the horizontal and is 100m above the level of the river. If the bank on the far side is 15 m below the top of the takeoff ramp, what minimum velocity must he have to reach the opposite bank?


Homework Equations


\Deltay = v0yt - .5gt2
\Deltax = v0xt
R = \frac{v<sup>2</sup>sin2\Theta}{g}

The Attempt at a Solution



I've made several attempts to solve this during the test, I'm completely stuck though.
With having 2 variables missing I can't solve the conventional way (Solving for t using the y displacement) because I don't know the 'y' component of velocity.

I don't see how it is possible to solving for t in the x dimension because, ini x velocity is unknown and time is unknown.

Using the max range formula wouldn't do me any good either, because there is a displacement in the y dimension. Even if I try to partially solve using the max range formula I fail, I do not know what distance before the cliff to use as range R
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Right now I'm thinking this problem isn't possible to solve, which is most likely not the case considering it was a test problem.

Extremely eager for someone to prove me wrong:
All help appreciated
-Pat
 
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I think you can do it. Just pick a letter like v for the initial speed and "pretend" you know it so you can write out the separate equations for the horizontal and vertical motion. Some people just can't handle an unknown v and have to run all that through with some number like 47 in place of the v, then erase all the 47's and put in v's afterward!

Anyway, I think you will end up with one horizontal equation and two vertical equations (velocity and distance) with unknowns v and t. You should then be able to solve two of the equations as a system to find v.
 
Let me know if anyone else has any other solutions.

I don't think that we had ever solved any problems as a system of equations. I tried doing this problem thoroughly as a substitution, but it got really messy.
 
You might post your work so we can see what went wrong.
Substitution is one method of solve a system of equations.
We aren't supposed to provide a solution; just help you with yours.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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