Coasting from rest down a hill, w/a variable slope, I reach a speed of 10 m/s at the halfway point?

Question

how fast would i go if i coasted down 2nd half of hill startng from rest?

Answer 1

If we ignore friction, the slope doesn't matter, only the height of the hill.
since
m*g*h/2=.5*m*100
we can calculate h as
h=100/g

using this result
m*g*h=m*100
and
m*100=.5*m*v^2
so
v=sqrt(200)

j

Answer 2

I'm a bit confused...starting from rest where? Are you starting from rest halfway down the slope; if you are, the velocity will be 10 mps like for the first half because the drop would be the same as for the first half.

But if you mean how fast will you be going when you reach the bottom of the slope when starting from the top, that's just a matter of recognizing all the potential energy at H is P(H) = mgH = 1/2 mv^2 = K(0) which is the kinetic energy at the bottom H = 0. H is the height at the top of the slope where you started from rest. This gives us v = sqrt(2gH); where v is your velocity at the bottom of the slope.

You can solve for v = sqrt(2gH) once you find H. And that you can get from mgH/2 = 1/2 mV^2; where V = 10 mps and H/2 is just half way down from the height H. That's all you fell from the rest point when reaching halfway down the slope. g = 9.81 m/sec^2 when on Earth's surface; so you have all the numbers for g and V. You can do the math.

This is a conservation of energy problem. The energy at H is E(H) = P(H) = mgH only before starting from rest. There is no kinetic energy K(H) = 0 at rest. At the bottom of the slope, because energy can be converted but not created or destroyed, we have E(H) = E(0) = K(0) = 1/2 mv^2 because P(0) = mg0 = 0 and there is no potential energy at the bottom of the slope. That is all the P(H) at H became K(0) at the bottom through conservation.