Question

A boy stands on a diving board and tosses a stone into a swimming pool. The stone is thrown from a height of $2.50m$ above the water surface with a velocity of $4.00m/s$ at an angle of ${60.0}^0$ above the horizontal. As the stone strikes the water surface, it immediately slows down to exactly half the speed it had when it struck the water and maintains that speed while in the water. After the stone enters the water, it moves in a straight line in the direction of the velocity it had when it struck the water. If the pool is $3.00m$ deep, how much time elapses between when the stone is thrown and when it strikes the bottom of the pool?

Answer 1

We first consider the vertical motion of the stone as it falls toward the water. The initial $y$ velocity component of the stone is
$v_{yi}=v_i\sin \theta =–\left(4.00m/s\right)\sin {60.0}^0=–3.46m/s$
and its $y$ coordinate is
$y_f=y_i+v_{yi}t+\frac{1}{2}{a}_y{t}^2=h+\left(v_i\sin \theta \right)t-\frac{1}{2}g{t}^2$
$y_f=2.50-3.46t-4.90t^2$
where $y$ is in $m$ and $t$ in $s$. We have taken the water’s surface to be at $y=0$. At the water,
$4.90t^2+3.46t–2.50=0$
Solving for the positive root of the equation, we get
$t=\frac{-3.46+\sqrt{(3.46{)}^2-4(4.90\left)\right(-2.50)}}{2\left(4.90\right)}$
$t=\frac{-3.46+7.81}{9.80}$
$t=0.443s$
The $y$ component of velocity of the stone when it reaches the water at this time $t$ is
$v_{yf}=v_{yi}+a_{y}t=–3.46–gt=–7.81m/s$
After the stone enters to water, its speed, and therefore the magnitude of each velocity component, is reduced by one-half. Thus, the $y$ component of the velocity of the stone in the water is
$v_{yi}=(–7.81m/s)/2=–3.91m/s$,
and this component remains constant until the stone reaches the bottom. As the stone moves through the water, its $y$ coordinate is
$y_f=y_i+v_{yi}t+\frac{1}{2}{a}_y{t}^2$
$y_f=-3.91t$
The stone reaches the bottom of the pool when $y_f=–3.00m$:
$y_f=–3.91t=–3.00\to t=0.767s$
The total time interval the stone takes to reach the bottom of the pool is
$\Delta t=0.443s+0.767s=1.21s$