A daredevil jumps a canyon 9.6 m wide. To do so, he drives a car up a 18 Solutio

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Explanation / Answer

let top of incline (height h) be the origin, and his launch speed be (v) at t=0 > hori > v cos 22 > vert > v sin 22 ---------------------------------- at time = t, his coordinates will be x = v cos 22 * t y = v sin 22 * t - 0.5 gt^2 at handing point, y = 0, t = time of flight (T) = hang time 0 = v sin 22 T - gT^2/2 T = 2v sin 22/g --------------------------------------… Range = gap cleared = 11.7 = v cos 22 * T = 2v^2 sin 22 cos 22/g 11.7 = v^2 sin 44/g v^2 = 9.6* 9.8/ sin 44 = 145.06 v (min) = 12.85 m/s =================== energy conservation> start1 = landing2 PE 1 + KE1 = PE 2 + KE2 > as he lands at same level from ground, PE1=PE2 so KE1 = KE2 v(min) = v(final) same speed landing

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