A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude $9.0m/s$. Find the motorcycle’s distance from the edge of the cliff after $0.5s$.

The correct option is A $\frac{\sqrt{349}}{4}m$
Draw a diagram of given situation.

Find the motorcycle’s distance from the edge of the cliff.

Given,

$u_x=9m/s,u_y=0$
$a_x=0,a_y=-g$

At $t=0.5s$

Horizontal distance,
$x=u_{x}t=\left(9.0m/s\right)\left(0.50s\right)=4.5m$

Vertical distance,
$y=-\frac{1}{2}g{t}^2=-\frac{1}{2}\left(10\right)(0.50{)}^2$

$-\frac{5}{4}m$

The negative value of $y$ shows that this time the motorcycle is below its starting point.The motorcycle's net distance from the origin at this time,

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