A basketball player shoots a free throw at a 50 degree angle. Assume the ball is released at a height of 1.8 meters, the hoop is 3 meters off of the floor and 4.6 meters away from the shooter. With what velocity should the player release the ball to hit nothing but net?
10.58 m/s
let speed of release = x
horizontal speed =u = x cos 50
initial vertical speed Vi = x sin 50
4.6 = (x cos 50) t so t = 4.6/(x cos 50)
3 = 1.8 + (x sin 50) t - 4.9 t^2
1.2 = (x sin 50)(4.6/x cos 50) - 4.9 (4.6^2)/(x^2 cos^2 50)
1.2 = 4.6 tan 50 - 4.9 (4.6^2)/(x^2 cos^2 50)
To find the velocity at which the player should release the ball to hit nothing but net, we can use the equations of motion and the principles of projectile motion.
1. First, let's break down the initial information given:
- The angle of release is 50 degrees.
- The initial height of the ball is 1.8 meters.
- The height of the hoop is 3 meters.
- The horizontal distance from the shooter to the hoop is 4.6 meters.
2. To solve this problem, we need to find the initial velocity of the ball (magnitude and direction) that will allow it to reach the hoop.
3. By analyzing the situation, we can determine that the vertical motion of the ball is influenced by gravity, while the horizontal motion is unaffected.
4. Let's analyze the vertical motion first:
- The initial vertical position (y0) is 1.8 meters.
- The final vertical position (y) is the height of the hoop, which is 3 meters.
- The initial vertical velocity (Vy0) is unknown.
- The final vertical velocity (Vy) will be zero at the peak height of the ball's trajectory.
- The acceleration due to gravity (g) is approximately 9.8 m/s^2, acting downward.
5. We can use the equation for vertical displacement in projectile motion to find the initial vertical velocity:
y = y0 + Vy0t - (1/2)gt^2
Substituting the given values:
3 = 1.8 + Vy0t - (1/2)(9.8)t^2
Simplifying the equation, we get:
0.5(9.8)t^2 - Vy0t - 1.2 = 0
6. Since we don't have values for time, we need to consider the apex of the projectile motion's trajectory when the vertical velocity becomes zero. At this point, the upward and downward motions also take the same amount of time.
7. Using the equation for the time of flight (time taken to reach the maximum height), we can find the time when Vy = 0:
Vy = Vy0 - gt => 0 = Vy0 - 9.8t => Vy0 = 9.8t
Substituting this value back into the equation from step 5:
0.5(9.8)(9.8t)^2 - (9.8t)t - 1.2 = 0
8. Now, we have a quadratic equation. By solving for t, we can find the time it takes for the ball to reach its peak height.
9. Let's move on to analyzing the horizontal motion:
- The horizontal distance from the shooter to the hoop is 4.6 meters.
- The horizontal initial velocity (Vx0) is unknown.
- The horizontal final velocity (Vx) will be the same as Vx0 since there is no horizontal acceleration.
10. We can use the equation for horizontal displacement in projectile motion to find the horizontal initial velocity:
x = x0 + Vx0t
Substituting the given values:
4.6 = 0 + Vx0t
Simplifying the equation, we get:
Vx0t = 4.6
11. Now we have a relationship between Vx0 and t. By combining this with the equation from step 8, we can solve for both Vx0 and t.
12. Once we find the values of Vx0 and t, we can calculate the initial velocity of the ball (V0). By using the Pythagorean theorem, we have:
V0 = √(Vx0^2 + Vy0^2)
13. Finally, we can solve for the required velocity at which the player should release the ball to hit nothing but net.
Note: The solution above provides a step-by-step explanation of how to approach and solve the problem. To obtain specific numerical values, you need to substitute appropriate values into the equations and solve them using algebraic methods.