From what height would a car..?

Question

From what height would a car have to be dropped in order to have the same kinetic energy that it would have when being driven at 1.00 x 108 km/hour?

I tried v*v/2g (since mgh = mv*v/2). Can;t see where I've made a mistake. Also converted km/hr to m/sec and used g=9.8m/s2

Answer 1

Potential energy at the top is = to the Kinetic energy at the bottom, so it's KE would equal it's UG.

KE=.5mv²

UG=mgh

so if it's KE at the bottom would equal (.5)(m)(108km/hr² ) therefor

g=9.8m/s²

108km/hr=30m/s

(.5)(m)(108km/hr² )=mgh

(.5)(30m/s² )=(9.8m/s²)h

h≅45

Answer 2

I got 45.87m. Be confident of your answer if you really cant find anything wrong.

Answer 3

KE = 1/2 mv^2

If it is dropped from height h, then KE on reaching the ground = mgh

Therefore, 1/2 mv^2 = mgh

Or, h = v^2/(2g)

v = 108 km/h = 108 * 1000/3600 m/s = 30 m/s

g = 9.8 m/s^2

h = 30^2/(2 * 9.8) m =  45.9 m

Ans: 45.9 m