Question

A person standing at the edge of a seaside
cliff kicks a stone over the edge with a speed
of 20 m/s. The cliff is 54 m above the water’s
surface, as shown.With what speed does the stone strike the
water?
Answer in units of m/s

Answer 1

As you can see from the drawing the Vox is 20m/s. So by knowing that Vox=20m/s you know that your Vfx = 20m/s because it is constant.

Therefore you already know the X component of the final velocity at which it strikes the water. Now you need to find the Y component.

\[Vy ^{2} = V _{oy}^{2} - 2g(y-y _{o})\]

Now you replace your known values to get your \[Vy\]
\[Vy ^{2} = 0 - 2(9.81m/s^{2})(-54m)\]
\[\sqrt{1059.48} =32m/s\]

Now you know the Vy component. Now to get the magnitude you use the formula
\[\sqrt{A _{x ^{2}}+{A _{y ^{2}}}}\]
\[\sqrt{(20 m/s){ ^{2}}+{(32.5{m/s) ^{2}}}}\]
=38.16m/s
That is the velocity at which it strikes the water.
If you have any questions I'll be here to answer.