A basketball player tries to make a half-court jump shot, releasing the ball at the height of the basket. Assuming that the ball is launched at 48.0°, 13.0 m from the basket, what speed must the player give the ball?
To find the speed the player must give the ball, we can use the principles of projectile motion. We can break down the launch velocity into its horizontal and vertical components. The horizontal component will determine how far the ball travels, while the vertical component will determine the height the ball reaches.
Given:
Angle of launch (θ) = 48.0°
Distance from the basket (d) = 13.0 m
1. Break down the launch velocity into its horizontal and vertical components.
- The horizontal component (Vx) remains constant throughout the flight and does not change due to gravity. It can be calculated using the equation Vx = V * cos(θ), where V is the initial velocity (speed of the ball) and θ is the angle of launch.
- The vertical component (Vy) changes due to gravity. It can be calculated using the equation Vy = V * sin(θ), where V is the initial velocity (speed of the ball) and θ is the angle of launch.
2. Find the time it takes for the ball to reach the halfway point.
- Since the ball is released at the height of the basket, it will take the same time to reach the halfway point as to reach the peak of its trajectory.
- The time (t) can be calculated using the equation t = Vy / g, where Vy is the vertical component of velocity and g is the acceleration due to gravity (9.8 m/s²).
3. Use the time calculated to find the horizontal distance traveled by the ball.
- The horizontal distance (Dx) can be calculated using the equation Dx = Vx * t, where Vx is the horizontal component of velocity and t is the time calculated in step 2.
4. Calculate the required initial velocity (speed of the ball).
- Since the ball needs to travel a horizontal distance of 13.0 m, the horizontal distance calculated in step 3 must be equal to 13.0 m.
- Use the equation Dx = 13.0 m and substitute the values of Vx and t from steps 1 and 2 to solve for V, the required initial velocity (speed of the ball).
By following these steps, you can determine the speed the basketball player must give the ball to make a half-court jump shot based on the given information.
To find the speed the player must give the ball, we can use the equations of motion for projectile motion.
Step 1: Dissect the given information:
- Launch angle (θ) = 48.0°
- Distance to the basket (range) = 13.0 m
Step 2: Break down the problem:
The motion of the basketball can be separated into horizontal and vertical components. The horizontal component is unaffected by gravity, and the vertical component is affected by gravity.
Step 3: Calculate the initial velocity of the ball:
To find the initial velocity, we need to calculate the horizontal and vertical components of the velocity (Vx and Vy).
The horizontal component (Vx) remains constant throughout the motion and can be calculated using the formula:
Vx = V * cos(θ)
The vertical component (Vy) depends on the launch angle and initial velocity and can be calculated using the formula:
Vy = V * sin(θ)
Since the ball is released at the height of the basket, the vertical displacement (Δy) is 0.
Since we are given the distance to the basket, which is the range (R), and the launch angle (θ), we can use the formula for the range of projectile motion:
R = (V^2 * sin(2θ)) / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Step 4: Solve for the initial velocity (V):
Rearranging the equation for range, we get:
V^2 = (R * g) / sin(2θ)
V = sqrt((R * g) / sin(2θ))
Substituting the given values:
V = sqrt((13.0 * 9.8) / sin(2 * 48.0°))
Calculating this expression gives us the required speed for the ball.