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.

 

[tomcue]

!

Hello, I see that you have attempted to solve the problem using the equations of kinematics. Your approach seems to be correct so far. However, in order to find the take off speed, you will need to use the equation v=u+at, where v is the final velocity, u is the initial velocity (take off speed), a is the acceleration (in this case, it is the acceleration due to gravity, which is -9.81 m/s^2), and t is the time.

To find the time, you can use the equation that you have already derived for the horizontal distance: 36/(xcos32) = t. Substituting this value of t into the equation for vertical distance, you can solve for the initial velocity (u).

v=u+at
0=u+(-9.81)(36/(xcos32))

Solving for u, we get:
u= 12.83 m/s

Therefore, the take off speed of the bike rider was 12.83 m/s. I hope this helps! Let me know if you have any further questions.