91Ó°ÊÓ

The given data can be listed below:

The width of the river is w = 48 m.

The distance of car above the river is y = 21.3 m .

The distance of car on other side of river is $y_0=1.8\mathrm{m}$.

The overall displacement of an object divided by the total time it takes to move that thing is its average velocity.

The distance traveled by the car is,

$$\begin{gathered} y-y_0=21.3\mathrm{m}-1.80\mathrm{m} \\ =19.5\mathrm{m} \end{gathered}$$

The time taken to cover the distance from the car is given by,

$$\begin{gathered} y-y_0=v_{0y}t+\frac{1}{2}{a}_y{t}^2 \\ t=\sqrt{\frac{2\left(y-y_0\right)}{a_y}} \end{gathered}$$

Here, $\left(y-y_0\right)$is the distance covered in vertical direction, $y_{0y}$is the initial velocity in vertical direction, $a_y$is the acceleration in vertical direction.

Substitute al the values in the above expression.

$$\begin{gathered} t=\sqrt{\frac{2\left(19.5\mathrm{m}\right)}{9.8\mathrm{m}/{\mathrm{s}}^2}} \\ =1.99\mathrm{s} \end{gathered}$$

The velocity of the car in the horizontal direction is given by,

$$\begin{gathered} \mathrm{x}-{\mathrm{x}}_0={\mathrm{v}}_{0\mathrm{x}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}{\mathrm{t}}^2 \\ \mathrm{x}-{\mathrm{x}}_0={\mathrm{v}}_{0\mathrm{x}}\mathrm{t} \\ {\mathrm{v}}_{0\mathrm{x}}=\frac{\mathrm{x}-{\mathrm{x}}_0}{\mathrm{t}} \end{gathered}$$

Here, $\mathrm{x}-{\mathrm{x}}_0$is the horizontal distance travelled by the car,$a_x$is the acceleration in x-direction whose value is zero.

Substitute all the values in the above,

$$\begin{gathered} v_0=\frac{48\mathrm{m}}{1.99s} \\ =24.1\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity of the car in the horizontal direction is 24.1 m/s

The velocity in the vertical direction is given by,

${\mathrm{v}}_{\mathrm{y}}={\mathrm{v}}_{0\mathrm{y}}+{\mathrm{a}}_{\mathrm{y}}\mathrm{t}$

Here,${\mathrm{v}}_{0\mathrm{y}}$is the initial velocity in the vertical direction,localid="1667623100755" ${\mathrm{a}}_{\mathrm{y}}$is the acceleration in vertical direction, and tis the time taken.

Substitute all the values in the above,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{y}}=\left(0\mathrm{m}/\mathrm{s}\right)+\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1.99\mathrm{s}\right) \\ =-19.55\mathrm{m}/\mathrm{s} \end{gathered}$$

The net velocity of the car is given by,

$\mathrm{v}=\sqrt{{\mathrm{v}}_{\mathrm{x}}^2+{\mathrm{v}}_{\mathrm{y}}^2}$

Substitute all the values in the above,

$$\begin{gathered} v=\sqrt{{\left(24.1\mathrm{m}/\mathrm{s}\right)}^2+{\left(-19.55\mathrm{m}/\mathrm{s}\right)}^2} \\ =\sqrt{961.086}\mathrm{m}/\mathrm{s} \\ =31\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the final velocity of the car is 31 m/s .