A 2.0 m-tall basketball player is standing on the floor 4.0 m from the basket. If he shoots the ball at a 52.0° angle with the horizontal, at what initial speed does the ball just barely goes through the rim neatly if the height of the basket is 3.05 m and the diameters of the basketball and the rim are 24 cm and 46 cm ?
Range = Vo^2*sin(2A)/g = 4.
Vo^2*sin(104)/9.8 = 4
Vo^2*0.0990 = 4
Vo = 6.4m/s[52o].
To find the initial speed of the basketball, we need to consider the motion of the ball as a projectile. First, let's determine the horizontal and vertical components of the initial velocity.
The horizontal component remains constant throughout the motion and can be found using the equation:
Vx = V * cos(θ)
where Vx is the horizontal component of the initial velocity, V is the initial velocity (what we need to find), and θ is the angle of the shot.
The vertical component of the initial velocity can be found using the equation:
Vy = V * sin(θ)
where Vy is the vertical component of the initial velocity.
To calculate the time of flight, we will use the equation:
t = (2 * Vy) / g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Since the basketball must clear the rim, the height reached by the ball should be greater than the height of the rim. We can calculate the height reached by the ball using the equation:
h = Vy * t + (1/2) * g * t²
where h is the height reached by the ball.
Now we can set up the following equation to solve for the initial velocity:
h = 3.05 m + d_ball/2 + d_rim/2 + 0.01 m
Here, d_ball is the diameter of the basketball (0.24 m) and d_rim is the diameter of the rim (0.46 m). The additional 0.01 m is added to allow for the ball to go through the net neatly.
Finally, we can substitute the values into the equation and solve for V.
At what point do you need help?