Question

You push a matchbox car horizontally down a track with multiple hills in it. You give the car an initial velocity of v1=6m/s. There are 4 hills in order of shortest to highest. (h1=.5m;h2=1.1m;h3=1.8m;h4=2m.) Which hills would you make it over? What will your velocity be at the top of hill 2 (1.1m?)

Answer 1

Kinetic energy is converted to potential energy when car rises up to the hill.

0.5 mv^2 = mgh , you can divide by mass.
Then h = 0.5v^2 / g
This is the maximum height the car can reach

Answer 2

Next question ( needed a piece of paper)

vf is speed on the top of hill , h = 1.1 m

Mechanical energy conserves:

0.5mv^2 = mgh + 0.5mvf^2

Divide again mass away

0.5 v^2 = gh + 0.5 vf^2

Multiply by 2, you get rid of 0.5

v^2 = 2gh + vf^2
Then
vf = square root ( v^ 2 - 2gh)

Note change of sign!

Try always solve first with symbols, then
Add values.