What Was the Take Off Speed of the Bike Rider?

[Paul_H]

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

 

[Delphi51]

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.