A basketball leaves a player's hands at a height of 2.20 m above the floor. The basket is 2.80 m above the floor. The player likes to shoot the ball at a 36 degree angle. If the shot is made from a horizontal distance of 9.00 m and must be accurate to -.22 +.22 m (horizontally), what is the range of initial speeds allowed to make the basket?

To determine the range of initial speeds allowed to make the basket, we can use the projectile motion equations.

First, let's find the time it takes for the basketball to reach the basket. We'll assume that the horizontal distance of 9.00 m is the range (R).

Step 1: Find the horizontal component of the initial velocity (Vx).
Vx = initial velocity * cos(angle)

Step 2: Find the vertical component of the initial velocity (Vy).
Vy = initial velocity * sin(angle)

Step 3: Find the time it takes for the basketball to reach the basket.
R = Vx * t
t = R / Vx

Step 4: Find the maximum height (H) reached by the basketball during its flight.
H = Vy * t - 0.5 * g * t^2

Step 5: Since the player wants the shot to be accurate within ± 0.22 m horizontally, we need to find the minimum and maximum initial speeds that result in the basketball going through the basket.

Let's calculate the range of initial speeds using the given values:

Given:
Height of player's hand above the floor (h1) = 2.20 m
Height of the basket above the floor (h2) = 2.80 m
Angle of the shot (θ) = 36 degrees
Horizontal distance to the basket (R) = 9.00 m
Accuracy range (Δx) = 0.22 m

Step 1: Calculate Vx and Vy using the given angle:
Vx = initial velocity * cos(angle)
Vy = initial velocity * sin(angle)

Step 2: Calculate the time to reach the basket:
t = R / Vx

Step 3: Calculate the maximum height:
H = Vy * t - 0.5 * g * t^2

Step 4: Calculate the minimum and maximum initial speeds:
Minimum speed (V_min) = sqrt((R + Δx) * g / (2 * sin(θ) * cos(θ)))
Maximum speed (V_max) = sqrt((R - Δx) * g / (2 * sin(θ) * cos(θ)))

Let's plug in the given values into the formulas:

Vx = initial velocity * cos(36 degrees)
Vy = initial velocity * sin(36 degrees)

t = 9.00 m / (initial velocity * cos(36 degrees))
H = initial velocity * sin(36 degrees) * (9.00 m / (initial velocity * cos(36 degrees))) - 0.5 * 9.8 m/s^2 * (9.00 m / (initial velocity * cos(36 degrees)))^2

We can now solve for the minimum and maximum initial speeds:

Minimum speed:
V_min = sqrt((9.00 m + 0.22 m) * 9.8 m/s^2 / (2 * sin(36 degrees) * cos(36 degrees)))

Maximum speed:
V_max = sqrt((9.00 m - 0.22 m) * 9.8 m/s^2 / (2 * sin(36 degrees) * cos(36 degrees)))

Calculating the values:

V_min = sqrt((9.22 m) * 9.8 m/s^2 / (2 * 0.58779 * 0.80902))
V_min = 9.76 m/s (rounded to two decimal places)

V_max = sqrt((8.78 m) * 9.8 m/s^2 / (2 * 0.58779 * 0.80902))
V_max = 10.08 m/s (rounded to two decimal places)

Therefore, the range of initial speeds allowed to make the basket is 9.76 m/s to 10.08 m/s.

To find the range of initial speeds allowed to make the basket, we can use the projectile motion equations.

The horizontal motion of the basketball is independent of its vertical motion. So, let's start by finding the horizontal distance the basketball will travel.

Given:
Initial height (h₀) = 2.20 m
Basket height (h) = 2.80 m
Horizontal distance (d) = 9.00 m

To find the horizontal distance, we can use the equation:
d = v₀ * t

Since the basketball is shot at an angle, its initial velocity (v₀) can be split into horizontal (v₀x) and vertical (v₀y) components.

v₀x = v₀ * cos(θ)
v₀y = v₀ * sin(θ)

where θ is the angle of projection which is 36 degrees.

Solving for t in terms of v₀y:
h = v₀y * t + (1/2) * g * t²

Here, g is the acceleration due to gravity, which is approximately 9.8 m/s².

Simplifying the equation:
2.80 = v₀ * sin(θ) * t - (1/2) * 9.8 * t²

Now, we can find t in terms of v₀y:
2.80 = v₀ * sin(θ) * t - 4.9 * t²

To find the time of flight (t), we can set the equation to zero:
4.9 * t² - v₀ * sin(θ) * t + 2.80 = 0

Now, we can solve this quadratic equation to find t.

Once we have t, we can now find the value of v₀x using the equation:
d = v₀x * t

Now, we can substitute the values of t, θ, and d to solve for v₀x.

Next, to find the range of initial speeds allowed, we can use the horizontal distance plus/minus the accuracy range, given as -0.22 m and +0.22 m.

Range of initial speeds = v₀x ± accuracy range

By substituting the values of v₀x and the accuracy range, we can find the range of initial speeds allowed to make the basket.