A 1.85 m tall basketball player attempts a goal 9.3 m from a basket (3.05 m high). If he shoots the ball at a 45° angle, at what initial speed must he throw the basketball so that it goes through the hoop without striking the backboard?
hf=hi+vi'*t+1/2 g t^2
where vi' is the vertical component of speed initially (Vsin45)
In the horizontal
d=VCos45*t
now t is the same in each equation, form the second, t=9.3*1.414/V; put that in the first equation for t..
3.05=1.85+V*.707*(9.3*1.414/V)+4.9(9.3*1.414/V)^2
multiply through both sides by V^2, then you have a quadratic, solve using the qudaratic equation.
To find the initial speed at which the basketball player must throw the ball, we can use the principles of projectile motion and solve the problem in two dimensions (horizontal and vertical).
1. First, let's break down the information given:
- The player's initial height (y0) is 1.85 m.
- The horizontal distance from the player to the basket (x) is 9.3 m.
- The height of the basket (yf) is 3.05 m.
- The launch angle (θ) is 45°.
2. Since the launch angle is 45°, we know that the vertical component of the initial velocity (V0y) will be equal to the horizontal component of the initial velocity (V0x). So, V0y = V0x.
3. Now, let's use the equations of motion for projectile motion. The horizontal distance (x) can be calculated using the equation:
x = V0x * t
Since the player wants the ball to go through the hoop without touching the backboard, we can assume the time of flight (t) will be the time it takes for the ball to reach its maximum height and then descend to the height of the hoop.
Therefore, t = 2 * t_max
Where t_max can be calculated using the equation:
V0y = g * t_max
(g is the acceleration due to gravity)
4. The vertical displacement (y) can also be calculated using the equation:
y = V0y * t - 0.5 * g * t²
Since the ball starts at a height of 1.85 m and ends at the height of the hoop (3.05 m above the ground), y becomes:
3.05 - 1.85 = V0y * t_max - 0.5 * g * (2 * t_max)²
5. Now, we substitute V0y from Step 3 into Step 4 equation to get rid of V0y:
3.05 - 1.85 = (g * t_max) * t_max - 0.5 * g * (2 * t_max)²
6. Rearrange the equation to solve for t_max:
0.5 * g * (2 * t_max)² - (g * t_max) * t_max + (3.05 - 1.85) = 0
7. Simplify the equation:
2 * (t_max)² - 2 * (t_max)² + 1.2 = 0
8. Solve the quadratic equation for t_max. In this case, the equation becomes linear because both terms are the same:
t_max = √(1.2 / 2) = √0.6 ≈ 0.7746 seconds
9. Since t = 2 * t_max, we find:
t ≈ 2 * 0.7746 ≈ 1.5492 seconds
10. Now that we have the total time of flight, we can calculate the horizontal velocity (V0x) using the equation:
x = V0x * t
9.3 = V0x * 1.5492
11. Rearrange the equation to solve for V0x:
V0x ≈ 9.3 / 1.5492 ≈ 6 m/s
Therefore, the basketball player must throw the ball with an initial speed of approximately 6 m/s to make the basket without touching the backboard.