Suppose a rocket-propelled motorcycle is fired from rest horizontally across a c

Question

Suppose a rocket-propelled motorcycle is fired from rest horizontally across a canyon 1.10 km wide.

(a) What minimum constant acceleration in the x-direction must be provided by the engines so the cycle crosses safely if the opposite side is 0.820 km lower than the starting point?____ m/s^2

(b) At what speed does the motorcycle land if it maintains this constant horizontal component of acceleration? Neglect air drag, but remember that gravity is still acting in the negative y-direction.______ m/s

Explanation / Answer

part a )

In the vertical direction y: where down is positive

s= distance = 0.820 km = 820 m

a= acceleration = gravity = 9.8 m/s^2

v=initial velocity = 0

s= 0.5 gt^2 + vt

t= (2s/g) = 12.94 s

in the horizontal direction x

s= distance = 1.1 km = 1100 m

v=initial velocity = 0

s= 0.5 at^2 + vt

a= 2s/t^2 = 13.14 m/s^2

part b )

velocity in x direction

vf = v + at

vf = 13.14 * 12.94 = 170.03 m/s

in y direction

vf = -gt

vf = 9.8 x 12.94 = -126.812 m/s

v = sqrt(vx^2 + vy^2)

v = sqrt(170.03 ^2 + (-126.812)^2)

v = 212.11 m/s

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