Question

The current world-record motorcycle jump is $\mathbf{77}\mathbf{.}\mathbf{0}\mathit{m}$, set by Jason Renie. Assume that he left the take-off ramp at$\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{°}$ to the horizontal and that the take-off and landing heights are the same. Neglecting air drag, determine his take-off speed.

Short answer

The take-off speed of the Jason Renie’ is $v_0=43.1\mathrm{m}/\mathrm{s}$

Step-by-step solution

Given information

Projection angle with horizontal is$∆={12}^0$

Range is given as $R=77.0m$

To understand the concept

This problem deals with the horizontal range of the particle which is nothing but the horizontal distance from the launch point at which the particle returns to the launch height. Using the formula for the range of the particle, take -off speed of the Jason Renie’s motorcycle can be found.

Formula:

The particle’s horizontal range can be given as,

$R=\frac{v_0^2}{g}\sin 2∆$ (i)

Where $v_0$is the initial velocity of the particle.

To find the take-off speed 

Using equation (i) the take-off speed can be written as,

$v_0=\sqrt{\frac{Rg}{\sin 2∆}}$

$v_0=\sqrt{\frac{\left(77.0m\right)\left(9.8m/s^2\right)}{\sin 2(12.0°)}}$

Thus, the take=off speed is $v_0=43.1\mathrm{m}/\mathrm{s}$