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

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

jamesj · Answer

Well, how much Kinetic Energy does the rock have when it strikes the water?

anonymous · Answer

conservation energy. K(kinetic energy)=U(potential energy) .  i think...

jamesj · Answer

Yes, so much much gravitational potential energy has been converted into KE. Use that to calculate the rock's speed.

shane_b · Answer

I'm not sure how conservation of energy helps here since you aren't given the mass.  I'd just use the kinematic equation:$$\large V_f^2=V_i^2+2ad$$

jamesj · Answer

And where, pray tell, does that equation come from? It comes directly from Conservation of Energy!

shane_b · Answer

As I said, you're not given a mass here.  If there's a way to solve it with COE I'm just not seeing :)

jamesj · Answer

The loss of potential energy of the rock is
PE = mgh, where h = 12m
The KE the rock has is (1/2)mv^2
This must be equal to the change in PE. Hence
$$ mgh = \frac{1}{2}mv^2 $$
and therefore
$$ v^2 = 2gh $$
Now solve for the speed, v

shane_b · Answer

Yea, as soon as I posted that I realized the masses cancel...dumb me.

jamesj · Answer

and by the way, this is where that kinematic equation comes from.

By the Work-Energy, theorem, the change in kinetic energy of an object is--barring other sorts of energy being involved--equal to the work done on it.

The work done is 
W = Fd = mad, 
F = force, d = distance, m = mass, a = acceleration

The change in KE is 
(1/2)mv^2 - (1/2)mu^2, 
where v = final velocity, u = initial velocity

Thus as W = change in KE

$$ mad = \frac{1}{2}m(v^2 - u^2) $$
Canceling mass terms and rearranging,
$$ v^2 = u^2 + 2ad $$