# Just as a car tops a 47-meter high hill with a speed of 36 km/h, it runs out of gas and coasts...

## Question:

Just as a car tops a 47-meter high hill with a speed of 36 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?

## Kinetic energy:

The term kinetic energy is considered only when the body possesses motion. The kinetic energy directly varies the velocity of the object. In S.I system, its measurable unit is Joules.

Given data:

• The initial height of the car is {eq}{h_1} = 47\,{\rm{m}} {/eq}
• The speed of the car on the hill is {eq}v = 36\,{\rm{km/h}} = \left( {\dfrac{{36 \times 1000}}{{3600}}} \right)\,{\rm{m/s}} {/eq}

As from the given data, to climb the next hill there will be only the potential energy exists.

The expression for the 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} &= 47 + \dfrac{1}{{2 \times 9.8}}\left( {\dfrac{{36 \times 1000}}{{3600}}} \right)\\ {h_2} &= 47 + 0.510\\ {h_2} &= 47.510\,{\rm{m}} \end{align*} {/eq}

Thus the height of the next hill is {eq}{h_2} = 47.510\,{\rm{m}} {/eq} 