A woman stands on a cliff 20 m high and throws a rock with speed 20 m/s at an angle of 30ø with the horizontal

Question

The magnitude of the velocity with which the rock hits the ground below is

Answer 1

Projectile motion

I'm assuming she's aiming down not up.

The x component of velocity is
20cos(30)=17.32 m/s

In the y-direction, the equation of motion is

y=20-20sin(30) t-1/2(9.8)t^2

solving for y=0 we get t=1.24 s

the speed at that time is
-20sin(30)*1.24-1/2(9.8)(1.24)^2
=-19.93 m/s

The magnitude is

sqrt(17.3^2+(-19.3)^2)=
sqrt(300+397.4)=26.4 m/s