A basketball player who is 2.0m tall is standing on the floor 10.0m from the basket,If he shoots the ball at a 40.0¤ angle witi the horizontal,at what initial speed must he throw so that it goes through the hoop without striking the backboard?The basket height is 3.05 m.

Question

A basketball player who is 2.0m tall is standing on the floor 10.0m from the basket,If he shoots the ball at a 40.0¤ angle witi the horizontal,at what initial speed must he throw so that it goes through the hoop without striking the backboard?The basket height is 3.05 m.

Answer 1

To solve this problem, we can use the principles of projectile motion. We need to calculate the initial speed at which the basketball player must throw the ball for it to go through the hoop without striking the backboard.

Step 1: Break down the initial velocity into horizontal and vertical components.

The initial velocity of the basketball can be divided into two components: one horizontal and one vertical. The vertical component determines the height the ball reaches, and the horizontal component determines the distance it travels towards the basket.

Step 2: Calculate the time it takes for the ball to reach the hoop.

We know that the horizontal distance from the player to the basket is 10.0m. We can use the formula:

horizontal distance = horizontal velocity * time

Since the initial horizontal velocity is equal to the horizontal component of the initial velocity, we can rewrite this equation as:

10.0m = initial horizontal velocity * time

Step 3: Calculate the height the ball reaches.

The height of the basket is given as 3.05m.
We can use trigonometry to calculate the vertical distance covered by the ball.
The vertical distance is given by:

vertical distance = initial vertical velocity * time + 0.5 * acceleration due to gravity * time^2

Since the ball goes through the hoop without striking the backboard, the vertical distance covered by the ball should be equal to the height of the basket. We can rewrite the equation as:

3.05m = initial vertical velocity * time + 0.5 * 9.8m/s^2 * time^2

Step 4: Calculate the initial velocity.

Since the initial velocity is at an angle relative to the horizontal, we can determine it using the horizontal and vertical components:

initial velocity = sqrt((initial horizontal velocity)^2 + (initial vertical velocity)^2)

Step 5: Solve the equations to find the initial velocity.

Now, we can combine the equations derived from steps 2, 3, and 4 to solve for the initial velocity.

First, solve equation (1) for time:
time = 10.0m / initial horizontal velocity

Then, substitute this value of time into equation (2):
3.05m = initial vertical velocity * (10.0m / initial horizontal velocity) + 0.5 * 9.8m/s^2 * (10.0m / initial horizontal velocity)^2

Simplify the equation and solve for initial vertical velocity:
3.05m = (10.0m * initial vertical velocity) / initial horizontal velocity + (0.5 * 9.8m/s^2 * (10.0m / initial horizontal velocity)^2

Finally, substitute the values from steps 1, 2, and 3 into equation (4) to find the initial velocity:
initial velocity = sqrt((initial horizontal velocity)^2 + (initial vertical velocity)^2)

Solve this equation to find the initial velocity in m/s, which is the speed at which the basketball player must throw the ball for it to go through the hoop without striking the backboard.

Answer 2

consider the vertical

hf=hi+Vsin40*t-4.9 t^2

consider the horrizontal
10=Vcos40 t
V=10/cos40t

put that in the first equation for V, then solve for t (notice it is a bit of algebra). have a pad of paper handy.

Then after you have t, V=10/tcos40