# A 2.00 METER TALL BASKETBALL PLAYER ATTEMPTS A GOAL 10.00 METERS FROM THE BASKET(3.05 METERS HIGH). IF HE SHOOTS THE BALL AT A 45.0 DEGREE ANGLE, AT WHAT INITIAL SPEED MUST HE THROW THE BALL SO THAT IS GOES THROUGH THE HOOP WITHOUT STRIKING THE BACKBOARD?

Initial hi=2.0

final hf=3.05

horizontal distance=10

Here are the equations:

hf=Hi + Vsin45*t - 4.9 t^2

and

10= Vcos45*t

In the second equation, solve for t in terms of the numbers and V. Then plug that into the first equation for t, and solve for V. It will be a quadratic, use the quadratic equation.

## hf=Hi + Vsin45*t - 4.9 t^2

(v sin 45)t = 1.05 + 4.905 t^2 ___ (1)

(v cos 45)t = 10 ___ (2)

(2)/(1):

10 tan 45 = 1.05 + 4.905 t^2

Solve for t, t = 1.35 s

Substitute t in (2),

(v cos 45)1.35 = 10

v = 10.47 m s^-1

## Why did the basketball player bring a ladder to the game? Because he wanted to reach new heights in his shooting skills!

To solve this problem, let's break it down step by step. Starting with the second equation:

10 = Vcos(45)*t

Since cos(45) = (√2)/2, we can simplify the equation to:

10 = (V√2)/2 * t

Now, let's solve for t:

t = 20 / (V√2)

Now, let's substitute this value of t into the first equation:

3.05 = 2 + Vsin(45) * t - 4.9t^2

Substituting the value of t:

3.05 = 2 + Vsin(45) * (20 / (V√2)) - 4.9 * (20 / (V√2))^2

Simplifying:

3.05 = 2 + (20 * V) / (V√2) - 4.9 * (20^2) / (V^2 * 2)

To avoid striking the backboard, let's find the initial speed at which the ball must be thrown. Solving this equation might require some serious math skills and a calculator. Good luck with that!

## To solve this problem, let's first calculate the time it takes for the ball to reach the hoop. We can use the second equation:

10 = V * cos(45) * t

Rearranging the equation to solve for t:

t = 10 / (V * cos(45))

Now, let's substitute this value of t into the first equation:

hf = hi + V * sin(45) * t - 4.9 * t^2

Substituting the values we know, with hi = 2.0 and hf = 3.05:

3.05 = 2.0 + V * sin(45) * (10 / (V * cos(45))) - 4.9 * (10 / (V * cos(45)))^2

Simplifying the equation:

3.05 - 2.0 = 10 * tan(45) - 4.9 * 10^2 / ((V * cos(45))^2)

1.05 = 10 - 4.9 * 100 / (V^2 * (cos(45))^2)

1.05 = 10 - 4.9 * 100 / (V^2 * 0.5^2)

1.05 = 10 - 196 / (V^2 * 0.25)

Re-arranging the equation to solve for V:

196 / (V^2 * 0.25) = 10 - 1.05

196 / (V^2 * 0.25) = 8.95

V^2 * 0.25 = 196 / 8.95

V^2 * 0.25 = 21.89

V^2 = 21.89 / 0.25

V^2 = 87.56

Taking the square root of both sides to solve for V:

V = sqrt(87.56)

V ≈ 9.357 m/s

Therefore, the initial speed at which the player must throw the ball is approximately 9.357 m/s.

## To solve this problem, we can use the equations of motion for projectile motion. The first equation relates the initial and final heights with the initial vertical velocity, time, and acceleration due to gravity:

hf = Hi + Vsinθ * t - (1/2) * g * t^2

In this equation:

- hf represents the final height of the basketball hoop (3.05 meters)

- Hi represents the initial height of the basketball player (2.00 meters)

- V represents the initial velocity of the basketball

- θ represents the launch angle (45 degrees)

- t represents the time taken for the basketball to reach the height of the hoop

- g represents the acceleration due to gravity (approximately 9.8 m/s^2)

The second equation relates the horizontal distance with the initial horizontal velocity and time:

d = Vcosθ * t

In this equation:

- d represents the horizontal distance to the basketball hoop (10 meters)

We can rearrange the second equation to solve for time (t):

t = d / (Vcosθ)

Substituting this value of t into the first equation, we get:

hf = Hi + (Vsinθ * d) / (Vcosθ) - (1/2) * g * (d / (Vcosθ))^2

Simplifying this equation further, we have:

hf = Hi + (d * tanθ) - (1/2) * g * (d / Vcosθ)^2

Now we can substitute the given values into the equation:

- Hi = 2.00 meters

- hf = 3.05 meters

- d = 10.00 meters

- θ = 45.0 degrees

- g = 9.8 m/s^2

3.05 = 2.00 + (10 * tan45) - (1/2) * 9.8 * (10 / Vcos45)^2

Simplifying the equation further, we have:

1.05 = 10 - (4.9 / V^2)

Rearranging the equation to isolate V^2, we get:

V^2 = 4.9 / (10 - 1.05)

V^2 = 0.54943

Finally, taking the square root of both sides to solve for V, we find:

V ≈ 0.7416 m/s

Therefore, the initial speed at which the basketball must be thrown to go through the hoop without striking the backboard is approximately 0.7416 m/s.