Question

A student drops a rock from a bridge to the
water 12 m below.
With what speed does the rock strike
the water?
Answer in units of m/s

Answer 1

v^2 = u^2 + 2as
a in this case is g = 9.81 or 10 depending what accuracy you want, and you know s, the distance. U is the starting velocity and I guess that this is zero.

Answer 2

Use the equation\[X(12m)=X(0)+V(original).time+gravity.{time^{2}}\]after cancelling, it becomes\[X(12m)=gravity.{time^{2}}\]solved for time\[Time(s)=\sqrt{X(12m)/gravity}\]then use the equation
\[V(final)=gravity.\sqrt{X(12m)/gravity}\]which becomes\[V(final)=\sqrt{X(12m)gravity}\]gravity constant is roughly rounded to 10meters/second square and all your units are already in metric system.

Answer 3

For those of you struggling with questions about things moving under constant accelerations - like gravity - the equations you need are very easy - but only if you do a few examples yourself.

For constant accelerations
v=u+a.t
s=u.t + 1/2 * a.t^2, and
v^2 = u^2 + 2.a.s

where u= initial velocity
v = final velocity
s = distance
a =acceleration
and t = time