Question

Question: A lead ball is dropped in a lake from a diving board $\mathbf{5}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{m}$ above the water. It hits the water with a certain velocity and then sinks to the bottom with this same constant velocity. It reaches the bottom $\mathbf{4}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{s}$after it is dropped. (a) How deep is the lake? What are the (b) magnitude and (c) direction (up or down) of the average velocity of the ball for the entire fall? Suppose that all the water is drained from the lake The ball is now thrown from the diving board so that it again reaches the bottom in$\mathbf{4}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{s}$ What are the(d) magnitude and (e) direction of the initial velocity of the ball?

Short answer

1. The lake is $38.08\mathrm{m}$deep.
2. Magnitude of average velocity is $9.02\mathrm{m}/\mathrm{s}$.
3. Direction of average velocity- the ball is going downward.
4. Magnitude of initial velocity is $14.5\mathrm{m}/\mathrm{s}$
5. Direction of initial velocity- the ball being thrown upward.

Step-by-step solution

Given information

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

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

To understand the concept

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 $∆\mathrm{y}$

(a): Calculations for depth of the lake

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)

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

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

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

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.

(b): Calculations for average velocity

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 $9.02\mathrm{s}$.

(c): Determination of the direction of average velocity

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

(d) and (e): Calculation for initial velocity of the ball and it’s direction

To find the initial velocity of the ball, we have

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

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