Question

The current world-record motorcycle jump is 77.0 m, set by Jason Renie. Assume that he left the take-off ramp at 13.0° to the horizontal and that the take-off and landing heights are the same. Neglecting air drag, determine his take-off speed.

Answer 1

Answer:

The take-off speed is 41.48 $\frac{m}{s}$

Explanation:

Given :

Range $R = 77$ m

Projectile angle $\alpha =$ 13°

From the formula of range,

  $R = \frac{v_{o} ^{2} }{g} \sin 2\alpha$

Find the velocity from above equation,

  $v_{o} = \sqrt{\frac{gR}{\sin 2 \alpha } }$

  $v_{o} = \sqrt{\frac{9.8 \times 77}{\sin 26} }$                              ( ∵ $g = 9.8$ $\frac{m}{s^{2} }$ )

  $v_{o} = 41.48$ $\frac{m}{s}$

Therefore, the take-off speed is 41.48 $\frac{m}{s}$