A basketball player attempting a free throw shoots the ball at an angle of 35 above the horizontal. The basket is 4.21 m from the foul line and 3.05 m above the floor. If the ball is released at a height of 1.83 m above the floor and it goes through the basket, what's the max height above the floor reached by the ball?
To determine the maximum height reached by the ball, we'll need to break down the motion of the basketball into horizontal and vertical components.
First, let's find the initial horizontal velocity of the ball. We know that the distance from the foul line to the basket is 4.21 m, and we can ignore any horizontal acceleration. Therefore, the time it takes for the ball to travel this distance horizontally is the same as the time it takes for the ball to reach the maximum height vertically.
We can use the formula for horizontal distance to calculate the time of flight:
horizontal distance = horizontal velocity × time.
Since the horizontal velocity remains constant throughout the trajectory, we can rewrite this formula as:
4.21 m = horizontal velocity × time.
Next, let's calculate the vertical acceleration due to gravity. We know that the ball is released at a height of 1.83 m, and it reaches a maximum height vertically. The vertical motion of the ball can be described using the following kinematic equation:
final vertical velocity² = initial vertical velocity² + 2 × vertical acceleration × vertical displacement.
Since the ball goes through the basket, the final vertical velocity is zero. We can also assume that the initial vertical velocity is zero (since the ball is released from rest). Therefore, we can rewrite the equation as:
0 = 0 + 2 × vertical acceleration × vertical displacement.
Simplifying the equation gives us:
vertical acceleration = -vertical displacement / 2.
Now, let's calculate the time taken for the ball to reach the maximum height. Since the initial vertical velocity is zero and the vertical acceleration is given by -vertical displacement / 2, we can use the following kinematic equation:
final vertical velocity = initial vertical velocity + vertical acceleration × time.
Since the final vertical velocity at the maximum height is zero, we can rewrite the equation as:
0 = 0 + (-vertical displacement / 2) × time.
This simplifies to:
time = 2 × vertical displacement / vertical acceleration.
Plugging in the values, we have:
time = 2 × 3.05 m / (-9.8 m/s²) = -0.62 s.
Note that the negative sign indicates that the ball is moving upwards.
Now, let's find the horizontal velocity. We know that time = 4.21 m / horizontal velocity. Plugging in the value for time, we can solve for horizontal velocity:
-0.62 s = 4.21 m / horizontal velocity.
Simplifying the equation gives us:
horizontal velocity = 4.21 m / (-0.62 s) = -6.79 m/s.
Note that the negative sign indicates that the ball is moving in the opposite direction of the positive x-axis.
Finally, let's calculate the maximum height above the floor reached by the ball. We can use the following kinematic equation for vertical displacement:
vertical displacement = initial vertical velocity × time + (1/2) × vertical acceleration × time².
Since the initial vertical velocity is zero and the vertical acceleration is given by -vertical displacement / 2, we can rewrite the equation as:
vertical displacement = (1/2) × (-vertical displacement / 2) × time².
This simplifies to:
4.21 m = (1/2) × (-vertical displacement / 2) × (-0.62 s)².
Simplifying the equation gives us:
vertical displacement = -(4.21 m) / (0.5 × 0.62 s²) = -13.51 m.
Note that the negative sign indicates that the ball is below the initial height.
To find the maximum height, we add the initial height to the vertical displacement:
max height = 1.83 m + (-13.51 m) = -11.68 m.
Thus, the maximum height above the floor reached by the ball is approximately 11.68 meters.
To find the maximum height reached by the ball, we will use the concept of projectile motion.
Given:
Angle above the horizontal: 35 degrees
Distance from the foul line to the basket: 4.21 m
Height of the basket above the floor: 3.05 m
Height at which the ball is released: 1.83 m
First, let's break down the initial velocity of the ball into its horizontal and vertical components.
Horizontal Component:
The horizontal component of the velocity remains constant throughout the motion and is given by:
Vx = V * cos(theta)
Vertical Component:
The vertical component of the velocity changes due to the effect of gravity. It can be calculated using the formula:
Vy = V * sin(theta)
Now, we can find the initial velocity (V) of the ball. Assuming gravity is 9.8 m/s^2, we can use the equation:
Vy = V * sin(theta)
Vy = V * sin(35)
V = Vy / sin(35)
Substituting the given values:
V = 1.83 / sin(35)
V ≈ 3.49 m/s
Next, we can determine the time (t) it takes for the ball to reach the basket using the horizontal component of the velocity:
Distance = Vx * time
4.21 = V * cos(theta) * t
Substituting the values we know:
4.21 = 3.49 * cos(35) * t
Simplifying, we find:
t ≈ 1.92 s
Now we can find the maximum height reached by the ball using the vertical component of the velocity:
H = Vy * t + (1/2) * g * t^2
Substituting the values we know:
H = 3.49 * sin(35) * 1.92 + (1/2) * 9.8 * 1.92^2
Simplifying, we find:
H ≈ 3.15 m
Therefore, the maximum height reached by the ball above the floor is approximately 3.15 meters.
The highest it could go if it must go in the basket would be when it is launched with enough velocity to clear the rim and bounce in off the backboard.
To do this properly, you need the diameter of the ball.
They must want you to make some approximations that let you treat the basketball as a point.
Assume that when the zero-radius ball reaches the basket it has the same height as the basket and a vertical velocity component of zero.