A dolphin jumps with an initial velocity of 13.0m/s at an angle of 41.0 ^\circ 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

consider the vertical velocity

Vf^2=Vi^2+2gh

solve for h. Vf (vertical velocity) is zero.

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To find the height above the water level at which the center of the hoop is positioned, we can use the concept of projectile motion. Here's how to solve it:

1. Resolve the initial velocity into its horizontal and vertical components. Given:
- Initial velocity (v0) = 13.0 m/s
- Angle (θ) = 41.0 degrees

The horizontal component of velocity (v0x) can be calculated using:
v0x = v0 * cos(θ)
v0x = 13.0 m/s * cos(41.0°)

The vertical component of velocity (v0y) can be calculated using:
v0y = v0 * sin(θ)
v0y = 13.0 m/s * sin(41.0°)

2. Determine the time (t) it takes for the dolphin to pass through the hoop. We can use the formula for vertical motion:
y = v0y * t + (1/2) * g * t^2
In this case, at the highest point of the motion, the vertical displacement (y) would be zero.

0 = v0y * t + (1/2) * g * t^2
Plug in the values:
0 = (13.0 m/s * sin(41.0°)) * t + (1/2) * 9.8 m/s^2 * t^2

Simplify the equation and solve for t:
0 = 6.85 t - 4.9t^2
4.9t^2 - 6.85t = 0
t(4.9t - 6.85) = 0

This equation gives two possibilities:
t = 0 (ignoring since it's the original position)
4.9t - 6.85 = 0
4.9t = 6.85
t ≈ 1.40 seconds

So, it takes approximately 1.40 seconds for the dolphin to pass through the hoop.

3. Using the time found in step 2, we can now calculate the height (h) of the hoop above the water level at that instant. We'll use the formula for vertical motion:
y = v0y * t + (1/2) * g * t^2

Plug in the values:
h = (13.0 m/s * sin(41.0°)) * 1.40 s + (1/2) * 9.8 m/s^2 * (1.40 s)^2
h = 6.85 m * 1.40 s + (1/2) * 9.8 m/s^2 * 1.96 s^2
h = 9.59 m + 9.61 m
h ≈ 19.2 m

The center of the hoop is approximately 19.2 meters above the water level.

Note: In this calculation, we have assumed that there is no air resistance and neglected any effects of buoyancy.