# Just as a car tops a 51 meter high hill with a speed of 84 km/h it runs out of gas and coasts...

## Question:

Just as a car tops a 51 meter high hill with a speed of 84 km/h it runs out of gas and coasts from there, without friction or drag. How high, to the nearest meter, will the car coast up the next hill?

## Potential energy:

The term potential energy is defined as the collected energy in the body or the object due to the position corresponding to the other objects. The potential energy of a body in vertical motion, directly varies with height.

Given data:

• The height of the hill is {eq}{h_1} = 51\,{\rm{m}} {/eq}
• The car with the speed is {eq}v = 84\,{\rm{km/h}} = \left( {\dfrac{{84 \times {{10}^3}}}{{3600}}} \right)\,{\rm{m/s}} {/eq}

The expression for the height of the hill is

{eq}e = k.e + p.e {/eq}

• Here {eq}k.e = \dfrac{1}{2}m{v^2} {/eq} is the kinetic energy.
• Here {eq}p.e = mg{h_1} {/eq} is the potential energy.

Substituting the values in the above equation as,

{eq}e = mgh + \dfrac{1}{2}m{v^2}\cdot\cdot\cdot{\rm(1)} {/eq}

The expression for the next hill height will be only the potential energy is

{eq}e = mg{h_2}\cdot\cdot\cdot{\rm(2)} {/eq}

Equating equation (1) and (2) as, by the law of conservation of energy is

{eq}\begin{align*} mg{h_2} &= mg{h_1} + \dfrac{1}{2}m{v^2}\\ {h_2} &= {h_1} + \dfrac{1}{{2g}}{v^2} \end{align*} {/eq}

Substituting the values in the above equation as,

{eq}\begin{align*} {h_2}& = {h_1} + \dfrac{1}{{2g}}{v^2}\\ {h_2} &= 51 + \dfrac{1}{{2 \times 9.8}}{\left( {\dfrac{{84 \times {{10}^3}}}{{3600}}} \right)^2}\\ {h_2} &= 78.77\,{\rm{m}} \end{align*} {/eq}

Thus the height of the next hill is {eq}{h_2} = 78.77\,{\rm{m}} {/eq}