A dolphin jumps with an initial velocity of 10.5 m/s at an angle of 32.0° 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 in meters?

Question

A dolphin jumps with an initial velocity of 10.5 m/s at an angle of 32.0° 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 in meters?

Answer 1

The time that the dolphin is in the air is 2*Vo sin32/g = 1.134 s. It rises for half of that time (0.567 s), and descends for the other half. The distance it rises is the average vertical velocity times the time going up or down, which is

H = (1/2)(Vo sin 32)*(Vo sin32/g)
(1/2)(Vo sin 32)^2/g

Answer 2

To determine the height of the center of the hoop above the water, we first need to break down the dolphin's initial velocity into horizontal and vertical components.

Given:
Initial velocity (v) = 10.5 m/s
Angle above the horizontal (θ) = 32.0°

The horizontal component of the velocity (v_x) can be found using the equation:
v_x = v * cos(θ)

The vertical component of the velocity (v_y) can be found using the equation:
v_y = v * sin(θ)

Now we can calculate v_x and v_y:

v_x = 10.5 m/s * cos(32.0°)
v_x ≈ 8.78 m/s

v_y = 10.5 m/s * sin(32.0°)
v_y ≈ 5.52 m/s

Since the dolphin is moving horizontally when it goes through the hoop, it means that the vertical component of its velocity (v_y) will be zero at that point. We can now solve for the time it takes for the dolphin to reach the center of the hoop.

Using the equation for vertical displacement:

Δy = v_y * t + (1/2) * a * t^2

Since the vertical displacement is zero (the dolphin starts and ends at the same height), we can simplify the equation to:

0 = v_y * t + (1/2) * a * t^2

Substituting v_y = 5.52 m/s and a = -9.8 m/s^2 (acceleration due to gravity) into the equation, we get:

0 = 5.52 m/s * t + (1/2) * (-9.8 m/s^2) * t^2

Simplifying further:

-4.9 m/s^2 * t^2 + 5.52 m/s * t = 0

We can now solve this quadratic equation to find the time it takes for the dolphin to reach the center of the hoop. Using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

a = -4.9 m/s^2
b = 5.52 m/s
c = 0

t = (-5.52 m/s ± √((5.52 m/s)^2 - 4 * -4.9 m/s^2 * 0)) / (2 * -4.9 m/s^2)

Now we can solve for t:

t = (-5.52 m/s ± √(30.4704 m^2/s^2)) / (-9.8 m/s^2)

Using the positive root (since time cannot be negative):

t = (-5.52 m/s + √(30.4704 m^2/s^2)) / (-9.8 m/s^2)
t ≈ 0.667 s

Now that we have the time (t) it takes for the dolphin to reach the center of the hoop, we can determine the height (h) by using the equation for vertical displacement:

h = v_y * t + (1/2) * a * t^2

Substituting v_y = 5.52 m/s, a = -9.8 m/s^2, and t ≈ 0.667 s into the equation:

h = 5.52 m/s * 0.667 s + (1/2) * (-9.8 m/s^2) * (0.667 s)^2
h ≈ 1.86 m

Therefore, the center of the hoop is approximately 1.86 meters above the water.

The time that the dolphin is in the air is 2*Vo sin32/g = 1.134 s. It rises for half of that time (0.567 s), and descends for the other half. The distance it rises is the average vertical velocity times the time going up or down, which is

yes

1.57

To determine the height of the center of the hoop above the water, we first need to break down the dolphin's initial velocity into horizontal and vertical components.