A basketball player throws the ball at a 60° angle above the horizontal to a hoop whose centre is located a horizontal distance L = 8.60 m from the point of release and at a height h = 0.80 m above it. What initial speed is required if the basketball is to reach the centre of the hoop? (Ignore the size of the ball.)

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To find the initial speed required for the basketball to reach the center of the hoop, we can use the equations of projectile motion.

First, we need to find the time it takes for the basketball to reach the horizontal distance L.

Using the horizontal motion equation:
L = V₀ * cos(theta) * t

where:
L = horizontal distance (8.60 m)
V₀ = initial velocity (what we want to find)
theta = angle above the horizontal (60°)
t = time

Rearranging the equation to solve for t:
t = L / (V₀ * cos(theta))

Next, we need to find the time it takes for the basketball to reach the vertical distance from the release point to the hoop, which is h = 0.80 m.

Using the vertical motion equation:
h = V₀ * sin(theta) * t - (1/2) * g * t^2

where:
h = vertical distance (0.80 m)
V₀ = initial velocity (what we want to find)
theta = angle above the horizontal (60°)
t = time
g = acceleration due to gravity (approximately 9.8 m/s^2)

Rearranging the equation to solve for t:
0 = V₀ * sin(theta) * t - (1/2) * g * t^2

Simplifying the equation:
(1/2) * g * t^2 = V₀ * sin(theta) * t

Dividing both sides of the equation by t (assuming t ≠ 0):
(1/2) * g * t = V₀ * sin(theta)

Now we have two equations with the same time variable t. We can equate them and solve for V₀.

L / (V₀ * cos(theta)) = (1/2) * g * t / (V₀ * sin(theta))

Canceling out t and rearranging the equation:
V₀ = L * g / (2 * sin(theta) * cos(theta))

Plugging in the known values:
L = 8.60 m
g = 9.8 m/s^2
theta = 60°

V₀ = 8.60 m * 9.8 m/s^2 / (2 * sin(60°) * cos(60°))

Simplifying:
V₀ = 34.77 m/s

Therefore, the initial speed required for the basketball to reach the center of the hoop is approximately 34.77 m/s.

To find the initial speed required for the basketball to reach the center of the hoop, we can use the equations of motion for projectiles.

First, let's break down the given information:
- The horizontal distance, L = 8.60 m
- The vertical height, h = 0.80 m
- The angle of projection, θ = 60°

We can start by analyzing the horizontal and vertical components of the initial velocity.

In the horizontal direction, there is no acceleration acting on the ball, so the velocity remains constant. Thus, we can use the equation:
Horizontal Distance (L) = Horizontal Velocity (V₀ * cos(θ)) * Time (t)

In this case, we want to find the time it takes for the ball to reach the hoop, so we can rearrange the equation:
Time (t) = Horizontal Distance (L) / Horizontal Velocity (V₀ * cos(θ))

Next, let's examine the vertical motion of the ball. Since there is a vertical acceleration due to gravity, we need to consider the equation for vertical displacement:
Vertical Height (h) = (Vertical Velocity (V₀ * sin(θ)) * Time (t)) - (0.5 * g * Time (t)^2)

Since the ball reaches the same height at the hoop as it started, the vertical displacement is zero. Therefore, we can set the equation equal to zero and solve for time:
0 = (V₀ * sin(θ)) * Time (t) - (0.5 * g * Time (t)^2)

Now, we have two equations with two unknowns (Velocity and Time). We can substitute the value of time from the first equation into the second equation, and solve for the velocity.

Let's plug in the values into the equations and solve mathematically:

Using the first equation:
T = L / (V₀ * cos(θ))

Substituting T into the second equation:
0 = (V₀ * sin(θ)) * (L / (V₀ * cos(θ))) - (0.5 * g * (L / (V₀ * cos(θ)))^2)

Simplifying:
0 = L * tan(θ) - (0.5 * g * L^2) / (V₀^2 * cos^2(θ))

Now, we can solve this equation for V₀, the initial velocity.

V₀^2 = (g * L^2) / (2 * h * sin^2(θ))

Taking the square root of both sides:
V₀ = √((g * L^2) / (2 * h * sin^2(θ)))

Plugging in the given values:
V₀ = √((9.8 m/s^2 * (8.60 m)^2) / (2 * 0.80 m * sin^2(60°)))

Calculating this expression will give us the initial speed (V₀) required for the basketball to reach the center of the hoop.