Question

A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude $9.0m{s}^{-1}$. Find the motorcycle's position, distance from the edge of the cliff and velocity after $0.5s$.

Answer 1

At $t=\mathrm{0..50}s$ the $x-$and $y$- coordinates are

$x=v_{0x}^t=\left(9.0m{s}^{-1}\right)\left(0.50s\right)$=$4.5m$

$y=-\frac{1}{2}g{t}^2=-\frac{1}{2}\left(10m{s}^{-2}\right)(0.50s{)}^2=-\frac{5}{4}m$

the negative value of $y$ shows that this time the motorcycles is below its starting point.

The motorcycle's distance from the origin at this time.

$r=\sqrt{x^2+y^2}=\sqrt{{\left(\frac{9}{2}\right)}^2+{\left(\frac{5}{4}\right)}^2}=\frac{\sqrt{349}}{4}m$

The components of velocity at this time are

$v_x=v_{0x}=9.0m{s}^{-1}$

$v_y=-gt=(-10m{s}^{-2})\left(0.50s\right)=-5m{s}^{-1}$

The speed (magnitude of the velocity ) at this time is

$v=\sqrt{v_x^2+v_y^2}$

$=\sqrt{(9.0m{s}^{-1}{)}^2+(-5m{s}^{-1{)}^2}}=\sqrt{106}m{s}^{-1}$