A basketball player throws the ball at a 53degree angle above the horizontal to a hoop which is located a horizontal distance L = 2.4 m from the point of release and at a height h = 0.9 m above it.What is the required speed if the basketball is to reach the hoop?

To calculate the required speed for the basketball to reach the hoop, we can use the laws of projectile motion.

Let's denote the initial velocity of the basketball as v0, the angle of projection as θ, the initial horizontal velocity as vx, and the initial vertical velocity as vy.

We can break down the initial velocity into its horizontal and vertical components:

vx = v0 * cos(θ)
vy = v0 * sin(θ)

At the highest point of the basketball's trajectory, the vertical velocity is 0, so we can determine the time it takes for the basketball to reach that point using the equation:

0 = vy - g * t
t = vy / g

where g is the acceleration due to gravity (9.8 m/s^2).

The horizontal distance traveled by the basketball can be calculated using the equation:

L = vx * t

Substituting the expressions for vx and t, we have:

L = v0 * cos(θ) * (vy / g)

Now, we can solve for v0:

v0 = L * g / (cos(θ) * vy)

We have the necessary values for L, g, θ, and vy. Plugging these values into the equation, we can calculate the required speed:

v0 = (2.4 m) * (9.8 m/s^2) / (cos(53 degrees) * sin(53 degrees))

Calculating this expression:

v0 = 4.715 m/s

Therefore, the required speed for the basketball to reach the hoop is approximately 4.715 m/s.

To find the required speed of the basketball, we will use the principles of projectile motion. We need to determine the initial velocity (speed) at which the basketball should be thrown in order to travel a horizontal distance of 2.4 m and reach a height of 0.9 m at a 53-degree angle above the horizontal.

Let's break down the problem into its horizontal and vertical components:

Horizontal Component:
The horizontal motion of the basketball is unaffected by gravity. Therefore, the horizontal component of the initial velocity remains constant throughout the motion.

Vertical Component:
The vertical motion of the basketball is influenced by gravity. We will use the equations of motion to analyze the vertical component.

Horizontal Component (Vx):
The horizontal component of velocity remains constant.
Vx = (initial velocity) * cos(angle)

Vertical Component (Vy):
The vertical component of velocity changes due to gravity.
Vy = (initial velocity) * sin(angle) - (gravitational acceleration * time)

Now, let's analyze the problem:

1. Calculate the time it takes for the basketball to reach its maximum height (h = 0.9 m). At the maximum height, the vertical velocity becomes zero.
Vy = 0
initial velocity * sin(angle) - (gravitational acceleration * time) = 0
time = (initial velocity * sin(angle)) / gravitational acceleration

2. Substitute the time obtained in step 1 into the equation for horizontal distance (L = 2.4 m):
L = (initial velocity * cos(angle)) * time

3. Rearrange the equation from step 2 to solve for the initial velocity:
initial velocity = L / (cos(angle) * time)

4. Substitute the values given in the problem:
angle = 53 degrees
L = 2.4 m
h = 0.9 m
gravitational acceleration = 9.8 m/s^2

5. Calculate the time using the equation derived in step 1:
time = (initial velocity * sin(angle)) / gravitational acceleration

6. Substitute the values obtained in step 5 and step 4 into the equation from step 3 to calculate the initial velocity:
initial velocity = L / (cos(angle) * [(initial velocity * sin(angle)) / gravitational acceleration])

7. Simplify the equation from step 6 by substituting known values:
initial velocity = (2.4 m) / (cos(53 degrees) * [(initial velocity * sin(53 degrees)) / 9.8 m/s^2])

8. Solve the equation from step 7 for the initial velocity:
Solve for initial velocity using algebraic techniques such as rearranging the equation or using a numerical method like iteration.

Following these steps, you should be able to determine the required speed at which the basketball should be thrown in order to reach the hoop.

h = 0.5g*t^2 = 0.9 m.

4.9*t^2 = 0.9
t^2 = 0.1837
Tf = 0.429 s. = Fall time.

Tr = Tf = 0.429s. = Rise time.

Y = Yo + g*Tr = 0
Yo = -g*Tr = 9.8 * 0.429 = 4.2 m/s.

Yo = Vo*Sin53 = 4.2
Vo = 4.2/Sin53 = 5.26 m/s. = The required speed.