What Was the Take Off Speed of the Bike Rider?

AI Thread Summary
The discussion focuses on calculating the take-off speed of a bike rider who leaves a ramp at a 32-degree angle and lands 36 meters away. The equations of motion are applied, with the horizontal distance and vertical motion analyzed separately. The user attempts to derive the time of flight and take-off speed using the equations S=ut+1/2at^2 for both horizontal and vertical components. They express their calculations and seek confirmation on their approach and the next steps. The conversation emphasizes the importance of correctly applying physics principles to solve the problem.
Paul_H
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1. A bike rider leaves a ramp at an angle of 32 degrees and lands at the same height after traveling a horizontal distance of 36m. What was his take off speed?



2. S=ut+1/2at^2



3. My attempt:
Vertical : s u v a=9.81 t=?
Horizontal: s=36m u v a=0 t=?

Horizontal:
S=ut+1/2at^2
36=(xcos32)t
36/(xcos32) = t

Vertical
S=ut+1/2at^2
xsin32/4.905 = t

36/xcos32 = xsin32/4.905

36*4.905 (xsin32)(xcos32)

Thats my best attempt, hopefully someone can tell me if I am going wrong and what is the next step in order to complete the question. Thanks
 
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36*4.905 = (xsin32)(xcos32) looks okay to me.
To solve for x, divide both sides by sin(32)*cos(32)
and take the square root of both sides.
 
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