Conservation of Energy Help (p2)

[kissafilipino]

Homework Statement

From what height would a compact car have to be dropped to have the same kinetic energy that it has when being driven at 105 km/h? Unless otherwise directed, assume that air resistance is negligible.

Answer: __ m
Velocity: 105 km/h -> 105000 m/s

Height: ?

Homework Equations

Conservation of energy equation: (thanks to user MillerGenuine)
U= potential energy
K= kinetic energy
Conservation of
energy => Kf + Uf = Ki + Ui

K = 1/2mv^2
U = mg(delta h)

The Attempt at a Solution

My attempt was somewhat he same trying to input in the equation, what height = the same kinetic energy of 115 k/m. So I said, gravitational potential energy = kinetic energy
U = K
mg(delta h) = (1/2)mv^2
Since mass is in both sides, it can be taken out.
g(delta h) = (1/2)v^2
(9.81)(delta h) = (1/2)105000^2
(9.81)(delta h) = (1/2)11025000000
(9.81)(delta h) = 5512500000
delta h = 561926605 m?

 

[Pengwuino]

Keep your units straight. Kilometers/hour is not SI units.

 

[kissafilipino]

Exactly, that I understood, which is why I changed it from Km to meters
105 km/h -> 105000 m/s and used it in the equation: mg(delta h) = (1/2)mv^2

 

[Pengwuino]

You changed km -> m but didn't correctly change hours -> seconds. 105km/h is about the speed a car is driven. Does your speed sound reasonable?