Answer to Question #96276 in Mechanics | Relativity for hugo

Question

A daredevil decides to jump a canyon. It's walls are equally high and 12 m apart. He takes off by driving a motorcycle up a short ramp sloped at an angle of 19°. What minimum speed must he have in order to clear the canyon?

Answer 1

Let us use notation $\theta = 19^{\circ}$, $L = 12 m$.

The horizontal distance, covered by the projectile aimed at an angle $\theta$ with initial speed $v_0$ is $S = \frac{v_0^2 \sin 2 \theta}{g}$. This distance must be minimum $L$ in order for a daredevil not to fall. Hence, from the previous equation, $\frac{v_0^2 \sin 2 \theta}{g} = L \Rightarrow v_0 = \sqrt{\frac{ g L}{ \sin 2 \theta}} \approx 13.83 \frac{m}{s}$.