Question

A lead ball is dropped in a lake from a diving board $5.20{\text{ }}m$ 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 $4.80s$ 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 $4.80s$ . What are the
(D) Magnitude and
(E) Direction of the initial velocity of the ball?

Answer 1

Hint
For solving this question, we need to use the kinematic equations of motion. We need to choose the equation containing only one unknown. Substituting the values given in the question will give the result.
$\Rightarrow v = u + at$
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow {v^2} - {u^2} = 2as$
$\Rightarrow v =$ final velocity, $u =$ initial velocity, $s =$ displacement, $a =$ acceleration, and $t =$ time.

Complete step by step answer
Let the depth of the lake be $d{\text{ }}m$
As the ball is dropped, so the initial velocity of the ball is zero i.e. $u = 0$
The height of the diving board, $h = 5.20m$
Acceleration of the ball $a = g = 9.8m/{s^2}$
Let the ball hit the lake with $v$ velocity.
Substituting these in the third equation of motion
$\Rightarrow {v^2} - {u^2} = 2as$
$\Rightarrow {v^2} - {0^2} = 2(9.8)(5.2)$
On multiplying
$\Rightarrow {v^2} = 101.92$
$\Rightarrow v = \sqrt {101.92}$
Calculating the square root, we get
$\Rightarrow v = 10.096m/s$
Let the time spent during this event be ${t_1}$
From the first equation of motion, we have
$\Rightarrow v = u + g{t_1}$
Substituting the values of $v$ , $u$ and $g$
$\Rightarrow 10.096 = 0 + 9.8{t_1}$
$\Rightarrow {t_1} = \dfrac{{10.096}}{{9.8}}$
On dividing, we get
$\Rightarrow {t_1} = 1.03s$
The total time to reach the bottom of the lake is $t = 4.80s$
$\therefore$ time for the ball to reach the bottom of the lake from the surface
$\Rightarrow {t_2} = t - {t_1}$
$\Rightarrow {t_2} = 4.80 - 1.03$
So, we have
$\Rightarrow {t_2} = 3.77s$
As the velocity of the ball becomes constant after hitting the water, its acceleration becomes zero.
The equation of motion for zero acceleration is given by
$\Rightarrow s = vt$
Substituting $v = 10.09m/s$ , $t = 3.77s$ , and $s = d$ , we get
$\Rightarrow d = 10.096 \times 3.77$
$\Rightarrow d = 38.06m$
Hence, the depth of the lake is $38.06m$
As we know, average velocity
$\Rightarrow {v_{avg}} = \dfrac{{{\text{Total displacement (x)}}}}{{{\text{Total time (t)}}}}$
The total displacement travelled by the ball, $x = h + d$
$\Rightarrow x = 5.2 + 38.06$
$\Rightarrow x = 43.06m$
The total time $t = 4.80s$
$\Rightarrow {v_{avg}} = \dfrac{{43.06}}{{4.8}}$
$\Rightarrow {v_{avg}} = 8.971m/s$
So, the average velocity is $8.971m/s$
As the direction of displacement is downwards, therefore the direction of average velocity is also downwards.
Now, as the water is drained, the ball’s motion will be uniformly accelerated. The motion will not be uniform along the depth of the water, as in the previous case. This is because; the water exerted the upward buoyant force on the ball equal to its weight which made its acceleration zero.
Let the ball be thrown with initial velocity $u{\text{ }}m/s$.
Now, the total displacement of the ball $=$ depth of the lake $=$ $38.06m$
Total time to reach the bottom of the lake $t = 4.80s$
From the second equation of motion, we have
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
Substituting $s = 38.06m,{\text{ }}a = g = 9.8m/{s^2},{\text{ }}t = 4.80s$
$\Rightarrow 38.06 = 4.8u + \dfrac{1}{2}(9.8){(4.8)^2}$
$\Rightarrow 38.06 = 4.8u + 112.896$
On rearranging
$\Rightarrow 4.8u = 38.06 - 112.896$
$\Rightarrow 4.8u = - 74.836$
Finally, $u = - 15.59m/s$
The negative sign indicates that the direction of initial velocity is upwards.
Hence, the magnitude of the initial velocity is $15.59m/s$ and its direction is upwards.

Note
The value $4.80s$ given in the question is the total time of the motion of the ball. Be careful not to use this value of time while calculating the depth of the lake. The calculation for the time required to cover the depth of the lake has to be made separately. It is a common mistake in these kinds of problems.