physics
posted by zacurea on .
The cart starts from rest at the top of a hill with height, H2. It rolls down the hill and at the bottom of the hill it has a speed of 10 m/s. Next it is at the top of same hill but pushed so that it starts at the top of the hill with a speed of 4.59 m/s. How fast is it going at the bottom of the hill?

Vf^2 = Vo^2 + 2g*h2.
Vf^2 =(4.59)^2 + 19.6*h2,
Vf^2 = 21.07 + 19.6h2,
Vf = sqrt(21.07 + 19.6h2).