91Ó°ÊÓ

Calculate the car's speed using the law of mechanical energy conservation. The law of mechanical energy conservation is written as follows:

$K_{\mathrm{f}}+U_{\mathrm{f}}=K_{\mathrm{i}}+U_{\mathrm{i}}$

We have, $K_{\mathrm{f}}$is the last kinetic energy, $K_{\mathrm{i}}$is the starting kinetic energy, $U_{\mathrm{f}}$is the last potential energy, and $U_i$is the starting potential energy.

The starting kinetic energy of the car is we have,

$K_{\mathrm{i}}=\frac{1}{2}m{v}_{\mathrm{i}}^2$

The last kinetic energy of the car we left with is,

$K_{\mathrm{f}}=\frac{1}{2}m{v}_{\mathrm{f}}^2$

We have, $m$is the mass of the car, $v_{\mathrm{f}}$is the last velocity, and $v_i$is the starting velocity of the car. The initial potential energy of the car is,

$U_{\mathrm{i}}=mg{y}_{\mathrm{i}}$

we have, $g$is the acceleration due to gravity with $g$ and $y_i$is the starting height.

The starting potential energy of the car we have,

$U_{\mathrm{f}}=mg{y}_{\mathrm{f}}$

we have, $y_{\mathrm{f}}$is the final height of the car.

Putting $\frac{1}{2}m{v}_{\mathrm{f}}^2$for $K_{\mathrm{f}},\frac{1}{2}m{v}_{\mathrm{i}}^2$for $K_{\mathrm{i}},mg{y}_{\mathrm{f}}$for $U_{\mathrm{f}}$, and $mg{y}_{\mathrm{i}}$for$U_{\mathrm{i}}$in the equation $K_{\mathrm{f}}+U_{\mathrm{f}}=K_{\mathrm{i}}+U_{\mathrm{i}}$and solve for $v_{\mathrm{f}}$.

$\frac{1}{2}m{v}_{\mathrm{f}}^2+mg{y}_{\mathrm{f}}=\frac{1}{2}m{v}_{\mathrm{i}}^2+mg{y}_{\mathrm{i}}$

$\frac{1}{2}m{v}_{\mathrm{f}}^2=\frac{1}{2}m{v}_{\mathrm{i}}^2+mg{y}_{\mathrm{i}}-mg{y}_{\mathrm{f}}$

$v_{\mathrm{f}}^2=v_{\mathrm{i}}^2+2g{y}_{\mathrm{i}}-2g{y}_{\mathrm{f}}$

$v_{\mathrm{f}}=\sqrt{v_{\mathrm{i}}^2+2g{y}_{\mathrm{i}}-2g{y}_{\mathrm{f}}}$

Putting $10\mathrm{m}/\mathrm{s}$for $v_{\mathrm{i}},9.80\mathrm{m}/{\mathrm{s}}^2$for $g,10\mathrm{m}$for $y_{\mathrm{i}}$, and $15\mathrm{m}$for $y_{\mathrm{f}}$.

$v_{\mathrm{f}}=\sqrt{(10\mathrm{m}/\mathrm{s}{)}^2+2\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)(10\mathrm{m})-2\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)(15\mathrm{m})}$

$=\sqrt{2}\mathrm{m}/\mathrm{s}$

$=1.4\mathrm{m}/\mathrm{s}$

As a result, on the other side of the valley, the car's speed is $1.4\mathrm{m}/\mathrm{s}$.