A1500 kg car is approaching a hill that has a height of 12 m. as the car reaches the bottom of the hill it runs out of gas and has a constant speed of 10 m/s. will the car make it to the top of the hill?

No, the car will not make it to the top of the hill.

Explanation:

Let ΔX be how long the slope of the hill is, Δx be how far the car will travel along the slope of the hill, Ф be the angle the slope of the hill makes with the horizontal(bottom of the hill), ki be the kinetic energy of the car at the bottom of the hill and vi be the velocity of the car at the bottom of the hill and kf be the kinetic energy of the car when it stop moving at vf.

Since Ф is the angle between the horizontal and the slope, the relationship between the angle and the slope and the height of the hill is given by

sinФ = 12/ΔX

Which gives you the slope as

ΔX = 12/sinФ

Therefore for the car to reach the top of the hill it will have to travel ΔX.

Ignoring friction the total work done is given by

W = ΔK

W = (kf - ki)

Since the car will come to a stop, kf = 0 J

W = -ki

m×g×sinФ×Δx = 1/2×m×vi^2

(9.8)×sinФ×Δx = 1/2×(10)^2

sinФΔx = 5.1

Δx = 5.1/sinФ

ΔX>>Δx Ф ∈ (0° , 90°)

(Note that the maximum angle Ф is 90° because the slope of a hill can never be greater ≥ 90° because that would then mean the car cannot travel uphill.)

Since the car can never travel the distance of the slope, it can never make it to the top of the hill.