Answer to Question #96276 in Mechanics | Relativity for hugo

Question #96276
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?
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Expert's answer
2019-10-10T09:53:49-0400

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

The horizontal distance, covered by the projectile aimed at an angle θ\theta with initial speed v0v_0 is S=v02sin2θgS = \frac{v_0^2 \sin 2 \theta}{g}. This distance must be minimum LL in order for a daredevil not to fall. Hence, from the previous equation, v02sin2θg=Lv0=gLsin2θ13.83ms\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}.


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