A basketball player throws the ball at a 36° angle above the horizontal to a hoop which is located a horizontal distance L = 8.2 m from the point of release and at a height h = 0.2 m above it. What is the required speed if the basketball is to reach the hoop?

Let t be the time of flight

Vcos32*t = 8.2
Vsin32*t -4.9 t^2 = 0.2

0.848 V*t = 8.2
0.530 V*t -4.9 t^2 = 0.2

First eliminate V by substitution and solve the resulting equation for t.

5.125 -4.9t^2 = 0.2
4.9t^2 = 4.925
t = 1.003 s
V = 8.2/(.848*1.003) = 9.64 m/s

To determine the required speed for the basketball to reach the hoop, we can use the equations of motion. The horizontal and vertical components of motion can be treated separately.

1. Vertical Component:
Let's consider the vertical motion of the basketball. The initial vertical velocity (Vy) at release is given by:
Vy = V * sin(angle)

where V is the initial speed and angle is the release angle (36°). Since the basketball reaches a maximum height of h = 0.2 m, we can use the equation:

h = (Vy^2) / (2 * g)

where g is the acceleration due to gravity (approximately 9.8 m/s^2). Rearranging the equation, we get:

Vy^2 = 2 * g * h

2. Horizontal Component:
The horizontal distance (L) traveled by the basketball can be calculated using the equation:

L = Vx * t

where Vx is the horizontal component of the initial velocity and t is the time of flight. Since there is no acceleration in the horizontal direction (assuming no air resistance), the horizontal component of the initial velocity (Vx) remains constant throughout the motion. It can be calculated using:

Vx = V * cos(angle)

The time of flight (t) can be determined using the equation:

t = L / Vx

3. Combining the Equations:
Now, we can substitute the values of t, Vx, and Vy in the equation L = Vx * t to eliminate t and find the required speed (V).

L = V * cos(angle) * (L / (V * cos(angle)))

Simplifying the equation, we have:

L = L

This shows that the value of L remains unchanged. Therefore, the required speed (V) needed for the basketball to reach the hoop is the same as the initial speed.

To summarize, the required speed for the basketball to reach the hoop is equal to the initial speed of the ball, which can be calculated using:
V = sqrt((2 * g * h) / sin(2 * angle))

To find the required speed, we can use the equations of motion and the principles of projectile motion. Let's break down the problem and find the solution step by step.

Step 1: Understand the given information.
- Angle of projection (θ) = 36° (angle above the horizontal)
- Horizontal distance (L) = 8.2 m
- Height (h) = 0.2 m

Step 2: Determine the components of initial velocity.
In projectile motion, the initial velocity can be broken down into two components: horizontal and vertical.

Horizontal component (Vx):
Since there is no acceleration in the horizontal direction, the velocity remains constant. Thus, the horizontal component of velocity (Vx) remains the same throughout the motion.

Vertical component (Vy):
The vertical component of velocity (Vy) is affected by gravity. For an object thrown upward or downward, the initial vertical velocity can be calculated using the formula:

Vy = V * sin(θ)

where V is the initial velocity and θ is the angle of projection.

Step 3: Determine the time of flight.
The time of flight (T) is the total time it takes for the object to reach the same vertical position from which it was released. It can be calculated using the formula:

T = (2 * Vy) / g

where g is the acceleration due to gravity.

Step 4: Determine the horizontal velocity (Vx).
The horizontal velocity (Vx) remains constant throughout the motion. The horizontal distance (L) can be calculated using the formula:

L = Vx * T

Step 5: Solve for the initial velocity (V).
To find V, we need to solve for Vy using the formula from Step 3:

T = (2 * Vy) / g

Rearranging, we get:

Vy = (g * T) / 2

Finally, we substitute the value of Vy into the equation for Vy from Step 2:

V = Vy / sin(θ)

Step 6: Plug in the values and calculate.
Using the given values,
- θ = 36°
- L = 8.2 m
- h = 0.2 m
- g = 9.8 m/s² (approximate value for acceleration due to gravity)

Calculate T:
T = (2 * Vy) / g = (2 * (g * T) / 2) / g = 2T

So, T = 2T which means T = 0.5 seconds.

Calculate Vx:
L = Vx * T
8.2 m = Vx * 0.5 s
Vx = (8.2 m) / (0.5 s) = 16.4 m/s

Calculate Vy:
Vy = (g * T) / 2 = (9.8 m/s² * 0.5 s) / 2 = 2.45 m/s

Finally, calculate the initial velocity (V):
V = Vy / sin(θ) = 2.45 m/s / sin(36°) = 4.24 m/s (rounded to two decimal places)

Therefore, the required speed for the basketball player to reach the hoop is approximately 4.24 m/s.

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