A stuntman is attempting to jump across a river on a motorcycle. The takeoff ram

Question

A stuntman is attempting to jump across a river on a motorcycle. The takeoff ramp is inclined at 53°, the river is 40.0 m wide, and the far bank of the river is 15.0 m lower than the top of the ramp. The river itself is 100.0 m below the top of the ramp. Neglect air resistance.
a.If the stuntman leaves the ramp with an initial speed of Vo=22m/s , will he land on the far bank of the river? Justify your response.
b. If the stuntman makes it across the river, what is his velocity at the instant he lands? If he does not make it across the river, what are the x- and y-coordinates of his landing point?

Explanation / Answer

= 53°, ramp is at origin, bank is at (40.0, -15.0) m, v0 = 22 m/s

x direction: x = v0cos*t, so t = x/(v0cos)

y direction: y = v0sin*t - gt2/2 = x*tan - gx2/(2v02cos2)

a) for y = -15.0 m,

-15.0 = 1.327x - 0.02798x2

solve it and get x = 56.86 m > 40.0 m, so he will land on the far bank.

b) the velocity = v

vx = v0cos = 13.24 m/s

vy = -[(v0sin)2 + 2*g*15.0] = -21.67 m/s (use vy2 = voy2 - 2gy)

magnitude of v = 25.4 m/s

tan-1(21.67/13.24) = 58.6o below horizontal

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