Question

Suppose a rocket-propelled motorcycle is fi red 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 component of acceleration? Neglect air drag, but remember that gravity is still acting in the
negative y-direction.

Answer 1

motion equation

$x=\frac{at^2}{2}$

$y=-\frac{gt^2}{2}$

Initial conditions

$g=9.8 \frac{m}{s^2}$

the cycle crosses safely if the...

$1000=\frac{at^2}{2}$

$-750=-\frac{gt^2}{2}$

From this we have acceleration

$a=\frac{1000\cdot 9.8}{750}=13.1 \frac{m}{s^2}$