Attempt to jump across a river on a motorcycle

  • Thread starter Chan M
  • Start date
  • #1
Chan M

Homework Statement



A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle. The takeoff ramp was inclined at 53 degrees, the river was 40 m. wide and the far bank was 15 m lower than the top of the ramp. The river itself was 100 m. below the ramp. You can ignore air resistance.What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? If his speed was only half the value found in (a), where did he land?

Xo = 0m
X = 40m
Ax = 0 m/s^2

Yo = 15m
Y= 0m
Ay = -9.8 m/s^2

2. Homework Equations


X = Xo + VoT + 0.5(Ax)T

Voy = Vo*sin(53)
Vox = Vo*cos(53)

X = Vox*T (since acceleration in the x-direction is zero)

3. The Attempt at a Solution


0 = 15 + VoyT - 4.9(Ay)T

0 = 15 + Vo * sin(53) * (40 / (Vo * cos(53) ) - 4.9 (40 / Vo * cos(53) )^2

Isolating Vo, I don't receive a valid answer.
My professor gave me the answer (17.8 m/s) but I still need to know how to solve the problem?

Thanks!
 
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Answers and Replies

  • #2
haruspex
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Isolating Vo, I don't receive a valid answer.
your equation is right, so it must be a mistake later. Please post your working (and please, not as an image).
 
  • #3
alejandromeira
In these problems it helps a lot to make a drawing to guide you.
Remember that in the first question, the landing point is 15m below the point of origin.
But in the second question, the point of contact is in the river, or on the cliff, you don't know that and you must calculate it. (you have inicial velocity)

Be careful with the signs of speeds, and accelerations, and applying the equations should not be too hard.
 

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