A basketball is thrown at 45° to the horizontal. The hoop is located 4.4 m away horizontally at a height of 1.1 m above the point of release. What is the required initial speed?

1.1 = Vi t - 4.9 t^2

t = 4.4/u
but u = Vi because tan 45 = 1

1.1 = u t - 4.9 t^2
1.1 = u t - 4.9 (4.4^2)/u^2

1.1 u^2 = u^2 (4.4) - 94.9

3.3 u^2 = 94.9
u = Vi = 5.36 m/s
sqrt(u^2+Vi^2) = 7.58 m/s

To find the required initial speed of the basketball, we can use the principles of projectile motion.

In projectile motion, an object moves in a curved path under the influence of gravity, with both horizontal and vertical components.

The horizontal component of the motion remains constant throughout the entire trajectory, while the vertical component accelerates due to gravity.

In this case, the horizontal distance traveled by the basketball (4.4 m) and the vertical displacement (1.1 m) are given. We need to find the initial speed (also known as the magnitude of the initial velocity).

To solve for the initial speed, we can break down the motion into horizontal and vertical components:

Horizontal Component:
The horizontal component of the velocity (Vx) can be found using the equation:
Vx = V * cos(theta)
where V is the initial speed and theta is the launch angle (45°).

Vertical Component:
The vertical component of the velocity (Vy) can be found using the equation:
Vy = V * sin(theta)

Since the time of flight (t) is the same for both components, we can use the formulas of motion to relate the distances traveled in the horizontal and vertical directions.

Horizontal Distance:
The horizontal distance (d) can be calculated using the equation:
d = Vx * t
where Vx is the horizontal component of the velocity.

Vertical Distance:
The vertical distance (h) can be calculated using the equation:
h = Vy * t - (0.5 * g * t^2)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Now we can solve for the initial velocity (V) by equating the horizontal and vertical distances to their given values:

For horizontal motion: d = 4.4 m
For vertical motion: h = 1.1 m

By substituting the equations for horizontal and vertical distances into the formulas for Vx and Vy, we can eliminate the variable t:

Vx * (d / Vx) = Vy * (h / Vy) - (0.5 * g * (h / Vy)^2)

Simplifying the equation gives:
d = h - (0.5 * g * (h / Vy)^2)

Now we can substitute the known values into the equation and solve for V:

4.4 = 1.1 - (0.5 * 9.8 * (1.1 / V * sin(45°))^2)

By rearranging and simplifying the equation, we can solve for V:

V^2 = (4.4 - 1.1) / (0.5 * 9.8 * (1.1 / sin(45°))^2)

V^2 = 3.3 / (0.5 * 9.8 * (1.1 / (sqrt(2)/2))^2)

V^2 = 3.3 / (0.5 * 9.8 * (1.1 / 0.7071)^2)

Solving for V gives:

V = sqrt( (3.3 * 2) / (0.5 * 9.8 * 1.1^2 * 0.7071^2) )

V ≈ 8.1 m/s

Therefore, the required initial speed for the basketball to reach the hoop is approximately 8.1 m/s.