A dolphin jumps with an initial velocity of 13.0at an angle of 44.0 above the horizontal. The dolphin passes through the center of a hoop before returning to the water. If the dolphin is moving horizontally when it goes through the hoop, how high above the water is the center of the hoop?

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To find the height above the water where the center of the hoop is located, we need to analyze the vertical motion of the dolphin.

Given:
Initial velocity (Vi) = 13.0 m/s
Launch angle (θ) = 44.0 degrees

First, we need to find the time it takes for the dolphin to reach the highest point of its trajectory. We can use the formula for vertical motion:

Vy = Vi * sin(θ)
where Vy is the vertical component of the velocity.

Using the given values, Vy = 13.0 m/s * sin(44.0) = 8.94 m/s.

Next, we use the equation for vertical motion to find the time of flight. At the highest point, the vertical component of velocity becomes zero. So, we can use the formula:

0 = Vy - g * t
where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time of flight.

Rearranging the equation, we get:
t = Vy / g = 8.94 m/s / 9.8 m/s² ≈ 0.912 seconds.

Now that we have the time of flight, we can calculate the height using the formula for vertical displacement:

h = Vi * sin(θ) * t - 0.5 * g * t²
where h is the height above the water.

Plugging in the values, we get:
h = 13.0 m/s * sin(44.0) * 0.912 s - 0.5 * 9.8 m/s² * (0.912 s)² ≈ 6.99 meters.

Therefore, the center of the hoop is approximately 6.99 meters above the water.