A copy-cat daredevil tries to reenact Evel Knievel’s 1974 attempt to jump the Sn

Question

A copy-cat daredevil tries to reenact Evel Knievel’s 1974 attempt to jump the Snake River Canyon in a rocket-powered motorcycle. The canyon is L = 398.1 m wide, with the opposite rims at the same height. The height of the launch ramp at one rim of the canyon is h = 8.55 m above the rim, and the angle of the end of the ramp is 45.0° with the horizontal.
Famous after his successful first jump, but still recovering from the injuries sustained in the crash caused by a strong bounce upon landing, the daredevil decides to jump again but to add a landing ramp with a slope that will match the angle of his velocity at landing. If the height of the landing ramp at the opposite rim is 3.21 m, what is the new required launch speed?

Explanation / Answer

Let vector a be the launch speed. Then the initial vertical and horizontal velocities are both (v2/2)a. v(t) = -9.8t + a*(v2/2) s(t) = -4.9t^2 + (v2/2)at + 6.23 Let t be the time in seconds from launch to landing. Then the horizontal distance traveled 382.3 = (v2/2)at And the vertical location s(t) = 0 = -4.9t^2 + (v2/2)at + 6.23 764.6/(av2) = t s(t) = 0 = -4.9(764.6/(av2))^2 + (v2/2)a(764.6/(av2)) + 6.23 a ? 60.7 m/s or 218.5 km/h For part b make s(t) = 3.19 instead of 0. And if you want to find the angle of the ramp for a smooth landing. Let vertical velocity = v, horizontal velocity = h and landing ramp angle = a. v(t) = -9.8t + a*(v2/2)

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