Physics Problem

Kayleigh..

A motor cyclist travelling at a speed of 40m/s rides straight over the edge of a 20m tall cliff. At the bottom of the cliff the ground falls away as a slope which makes an angle of 10 degrees to the horizontal.

Write an Octave programme which shows the (x,y) trajectort of the motorcyclist and the sloping ground. From the graph determine as accurately as you can, the position at which the cyclist hits the ground.

I can't seem to wrap my head around this one, for some reason the 10 degrees has me stumped, probably over complicating it but please help :P

Morbert

A motor cyclist travelling at a speed of 40m/s rides straight over the edge of a 20m tall cliff.

This means your parametric equation r(t) is

r(t) = (x(t),y(t)) = (40*t,-0.5*g*t^2)

(where g is 9.8 m/s^2).

At the bottom of the cliff the ground falls away as a slope which makes an angle of 10 degrees to the horizontal.

This means the ground can be represented by the line y = -20 - x*sin(10 degrees) I.e.

r'(t) = (t,-20-t*sin(pi/18))

In octave, you can plot these curves with the commands

hold on
ezplot(@(t) 40*t,@(t) -0.5*9.8*t.^2,[0,3])
ezplot(@(t) t,@(t) -20-t*sin(pi/18),[0,150])

[Edit] - Note that in the above case I am taking the 0-height mark to be the cliff top. If you want the cliff bottom to be the 0-height mark, just shift the 20m. I.e.

hold on
ezplot(@(t) 40*t,@(t) 20-0.5*9.8*t.^2,[0,3])
ezplot(@(t) t,@(t) -t*sin(pi/18),[0,150])