Question

Suppose a rocket-propelled motorcycle is fired from rest horizontally across a canyon 1.30 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 0.660 km lower than the starting point?
m/s²

(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

Answer 1

(a) First of all we need to find the time taken for the rocket to travel a vertical diatnce of 0.66 km under gravity. For this we use the kinematic equation,

Here the total vertical displacement is y = -0.66 km = -660 m

vertical component of initial velocity is

vertical component of acceleration

then

then

t = 11.605 s

Now we find the required horizontal component of acceleration using the equation

Here the horizontal distance x = 1.3km = 1300m

Horizontal component of initial velocity is zero

then the required horizontal accceleration is

                                                                 ax = 2*1300 / 11.60^2

                                                                ax = 19.30 m/s^2                     

(b) Now the horizontal component of final velocity is

                            

                                    = 0 + (19.3 m/s^2)(11.605 s)

                                    = 224.01 m/s

       And the vertical component of final velocity is

                               

                                    = 0 + (-9.81m/s^2)(11.605s)

                                    = -113.845 m/s

Then the final speed of the rocket is v = sqrt[(224.01 m/s)^2 + (-113.845 m/s)^2]

                                                         = 251.28 m/s