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A daredevil on a motorcycle leaves the end of a ramp with a speed of 37.0 m/s as in the figure below. If his speed is 35.1 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance. m

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Let:
? = the initial angle of elevation of the ramp

U = the initial speed = 37m/s
Uh = the horizontal component of U
Uv = the vertical component of U

(1) Uh = U * cos(?)

Because there are no frictional or air resistance losses, this horizontal speed will be maintained until the motorcycle is stopped. In particular, this will also be its horizontal speed at its highest point.

At its highest point, Vv = 0 and its speed is 35.1m/s;
therefore Vh = 35.1m/s and also

(2) Uh = 35.1m/s

Substituting into (1), we get
(3) 35.1 = 34 * cos(?)

Solving for ?,
(4) ? = acos(35.1 / 37) = 18.441225748995deg

Now,
(5) Uv = U * sin(?)
= 37 * sin(18.441225748995)
= 11.704272724102m/s

Max height is given by (ref 1, using equation (5) with v^2 = 0):

(6) h = Uv^2 / 2g
= 11.704272724102^2 / (2 * 9.81)

= 6.9821610601507672m above the end of the ramp.
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