Question

A basketball player shoots toward a basket 6.1 m away and 3.0 m above the floor. If the ball is released 1.8 m above the floor at an angle of 60° above the horizontal, what must the initial speed be if it were to go through the basket?

a) 7.0 m/s
b) 10.0 m/s
c) 14.0 m/s
d) 20.0 m/s

Answer 1

To find the initial speed needed for the ball to go through the basket, we can break down the initial velocity into its horizontal and vertical components. By using the equations of projectile motion and substituting the given values, we find that the initial speed must be approximately 10.0 m/s. So, the correct answer is b) 10.0 m/s.

Step-by-step explanation:

To solve this problem, we can break down the initial velocity into its horizontal and vertical components, and use the equations of projectile motion to find the initial speed. The horizontal velocity remains constant throughout the motion. Therefore, the horizontal component of the initial velocity is equal to the horizontal component of the final velocity, which is equal to the distance to the basket divided by the time of flight. The vertical component of the initial velocity can be calculated using the equation:

vy = vi * sin(θ)

where vy is the vertical component of the initial velocity, vi is the initial speed, and θ is the angle of release.

At the highest point of the trajectory, the vertical velocity is zero. Therefore, we can use the equation:

vy = vi * sin(θ) - g * t

where g is the acceleration due to gravity and t is the time of flight. Setting this equation equal to zero, we can solve for the time of flight:

t = vi * sin(θ) / g

Substituting this expression for t into the equation for the horizontal component of the velocity, we can solve for the initial speed:

vx = vi * cos(θ)

vi = d / (t * cos(θ))

where d is the distance to the basket. Plugging in the given values, we can calculate the initial speed:

vi = 6.1 / (1.8 * cos(60°))

vi ≈ 10.0 m/s

Thus, the correct answer is b) 10.0 m/s.