91Ó°ÊÓ

$\mathrm{h}=5.20\mathrm{m}$

Total time (T)$=4.80\mathrm{m}/\mathrm{s}$

The problem deals with the kinematic equation of motion. Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence the motion. The equations which are used in the study are known as kinematic equations of motion.

Formula:

The final velocity in the kinematic equation is expressed as,

$v_f^2=v_0^2+2ax$

The displacement is given by,

$∆\mathrm{x}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$

Here $∆\mathrm{x}$ can be replaced by $∆\mathrm{y}$

Here, first we have to find depth of lake by using total time of ball as follows

The speed of ball as it reaches the surface is given by$\left({\mathrm{v}}_{\mathrm{f}}\right)$

Consider $a=g$ and $x=h$Using equation (i)

$\begin{array}{rcl}{\mathrm{v}}_{\mathrm{f}}^2& =& 0+2\mathrm{gh}\\ {\mathrm{v}}_{\mathrm{f}}& =& \sqrt{2×9.81×5.20}\\ & =& 10.1\mathrm{m}/\mathrm{s}\end{array}$

The time for ball to fall from the board to the Lake Surface can be found as

localid="1662968094474" $∆\mathrm{y}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$

localid="1662968106380" $\mathrm{y}=\mathrm{h}\mathrm{and}{\mathrm{v}}_0=0$

$$\begin{gathered} \mathrm{h}=\frac{1}{2}{\mathrm{at}}^2 \\ {\mathrm{t}}_1=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}} \\ {\mathrm{t}}_1=\sqrt{\frac{2×5.20}{9.8}} \\ {\mathrm{t}}_1=1.029\mathrm{s} \end{gathered}$$

It would take $1.029\mathrm{s}$ for the ball to reach to the lake surface.

The time the ball spends descending in the lake is given by $t_2$ as follows

Let’s assume that D is the depth of the lake, so we have

$$\begin{gathered} {\mathrm{t}}_2=\frac{\mathrm{D}}{\mathrm{V}} \\ {\mathrm{t}}_2=\frac{\mathrm{D}}{10.01} \end{gathered}$$

So the total time

localid="1662968515316" $$\begin{gathered} \mathrm{T}={\mathrm{t}}_1+{\mathrm{t}}_2 \\ 4.80=1.029+\frac{\mathrm{D}}{10.01} \\ \mathrm{D}=38.08\mathrm{m} \end{gathered}$$

So, lake is $38.08\mathrm{m}$deep.

Average velocity can be found as

$$\begin{gathered} {\mathrm{v}}_{\mathrm{avg}}=\frac{\mathrm{D}+\mathrm{h}}{\mathrm{T}} \\ {\mathrm{v}}_{\mathrm{avg}}=\frac{38.08+5.20}{4.80} \\ {\mathrm{v}}_{\mathrm{avg}}=9.02\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the average velocity is $9.02\mathrm{s}$.

As ${\mathrm{v}}_{\mathrm{avg}}$ is positive, it is in the downward direction

To find the initial velocity of the ball, we have

$$\begin{gathered} ∆\mathrm{y}={\mathrm{v}}_0×\mathrm{t}+\frac{1}{2}×\mathrm{a}×{\mathrm{t}}^2 \\ \mathrm{h}+\mathrm{D}={\mathrm{v}}_0×\mathrm{T}+\frac{1}{2}×\mathrm{a}×{\mathrm{T}}^2 \\ 43.3={\mathrm{v}}_0×4.80+\frac{1}{2}×9.8×4.{80}^2 \\ {\mathrm{v}}_0=-14.5\mathrm{m}/\mathrm{s} \end{gathered}$$

As the initial velocity is negative and having magnitude $14.5\mathrm{m}/\mathrm{s}$, it is directed upward..