A trained dolphin leaps from the water with an initial speed of 12 m/s. It jumps directly towards a ball held by the trainer a horizontal distance of 5.50 m away and a vertical distance of 4.10 m above the water. In the absence of gravity the dolphin would move in a straight line to the ball and catch it, but because of gravity the dolphin follows a parabolic path well below the ball’s initial position. If the trainer releases the ball the instant the dolphin leaves the water, show that the dolphin and the falling ball meet.

To show that the dolphin and the falling ball meet, we will calculate the time it takes for the dolphin to reach the ball's horizontal position and the distance the ball falls vertically during that time.

First, let's find the time it takes for the dolphin to reach the ball's horizontal position:

We can use the kinematic equation:

\(s = ut + \frac{1}{2}at^2\)

where:
s = distance
u = initial speed
t = time
a = acceleration

Since the dolphin and the ball are moving horizontally, there is no horizontal acceleration. Therefore, the horizontal component of the dolphin's velocity remains constant at 12 m/s.

The horizontal distance traveled by the dolphin, \(s\), is 5.50 m. The initial horizontal velocity, \(u\), is 12 m/s, and the horizontal acceleration, \(a\), is 0 m/s^2.

So we have:

\(s = ut\)
\(5.50 = 12t\)

Solving for \(t\):

\(t = \frac{5.50}{12}\)
\(t \approx 0.4583\) seconds

Now let's calculate the distance the ball falls vertically during that time:

The vertical position of the ball can be determined using the free fall equation:

\(s = ut + \frac{1}{2}gt^2\)

where:
s = distance
u = initial vertical velocity
t = time
g = acceleration due to gravity

The vertical distance traveled by the ball, \(s\), is 4.10 m. The initial vertical velocity, \(u\), is 0 m/s (since the ball is released when the dolphin leaves the water), and the acceleration due to gravity, \(g\), is approximately 9.8 m/s^2.

So we have:

\(s = \frac{1}{2}gt^2\)
\(4.10 = \frac{1}{2} \cdot 9.8 \cdot t^2\)

Solving for \(t\):

\(t^2 = \frac{4.10}{\frac{1}{2} \cdot 9.8}\)
\(t^2 \approx 0.84\)
\(t \approx \sqrt{0.84}\)
\(t \approx 0.917\) seconds

Since the time taken for the dolphin to reach the ball's horizontal position is less than the time the ball takes to fall vertically, it shows that the dolphin and the falling ball meet.

To show that the dolphin and the falling ball meet, we need to determine if their paths intersect. We'll start by analyzing the motion of both the dolphin and the ball.

Let's consider the horizontal and vertical components of motion separately.

For the dolphin:
- Initial horizontal velocity (Vox) = 12 m/s (given)
- Vertical displacement (Δy) = -4.10 m (negative since it is above the water level)
- Horizontal displacement (Δx) = 5.50 m

For the ball:
- Initial horizontal velocity (Vox) = 0 m/s (since it is dropped by the trainer)
- Vertical displacement (Δy) = -4.10 m (same as the dolphin's vertical displacement)
- Horizontal displacement (Δx) = 5.50 m

Since both the dolphin and the ball have the same horizontal displacement, we need to determine if their vertical displacements are the same at some point during their motion.

We can use the equations of motion to determine the height reached by the dolphin and the time it takes.

For the dolphin:
Vertical displacement (Δy) = Voyo * t + (½) * g * t^2
where Voyo is the initial vertical velocity of the dolphin (0 m/s since it jumps straight up) and t is the time taken.

For the ball:
Vertical displacement (Δy) = 0.5 * g * t_ball^2
where t_ball is the time taken for the ball to fall.

Since both the dolphin and the ball have the same vertical displacement, we can equate the two equations:
Voyo * t + (½) * g * t^2 = 0.5 * g * t_ball^2

Now, let's solve this equation. We know the initial velocity of the dolphin is 12 m/s, and the acceleration due to gravity (g) is approximately 9.8 m/s^2.

0 * t + (½) * 9.8 * t^2 = 0.5 * 9.8 * t_ball^2
0 + 4.9 * t^2 = 4.9 * t_ball^2
t^2 = t_ball^2

As we can see, the time squared for both the dolphin and the ball are the same. This implies that their time of flight is equal. Since both the dolphin and the ball were released at the same instant, they meet at some point during their motion.

Therefore, the dolphin and the falling ball intersect in their paths.

The angle that dolphin leaves the water:

tanα = h/d =4.10/5.50 = 0.745,
α =arctan0.745 = 36.7º.
The x- and y- components of the initial velocity:
v(ox) = v(o) •cosα = 12•cos 36.7º = 9.62 m/s,
v(oy) = v(o) •sin α = 12• sin 36.7º = 7.17 m/s,
v(x)=v(ox)
x = v(x) •t,
t = x/ v(x) = 5.5/9.62 = 0.572 s.
y = v(oy) •t - g•t²/2 = 7.17•0.572 – 9.8•(0.572)²/2 = 2.5 m.
The ball’s location at this instant of time:
y = y(o) +v(o)t - g•t²/2 =
= 4.1 + 0•0.572 - 9.8•(0.572)²/2 = 2.5 m/s.