Question

Could you please draw the image and solve it

EXERCISE 3.6 Suppose a rocket-propelled motorcycle is fired from rest horizontally across a canyon 1.00 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 0.750 km lower than the starting point? (b) At what speed does the motorcycle land if it maintains this constant horizontal com- pon ent of acceleration? Neglect air drag, but remember that gravity is still acting in the negative y-direction.

Answer 1

In the vertical direction y: where down is positive

s= distance = 0.750 km = 750 m

a= acceleration = gravity = 9.81 m/s^2

v=initial velocity = 0

s= 0.5 gt^2 + vt

t= √(2s/g) = 12.37 s

in the horizontal direction x

s= distance = 1 km = 1000 m

v=initial velocity = 0

s= 0.5 gt^2 + vt

a= 2s/t^2 = 13.08 m/s