Mike wants to make a free throw; he realizes that if the ball is exactly 20cm above the hoop as it hits the backboard at the top of its trajectory, it will go in. The free throw line is 4.5 m from the plane of the backboard and the hoop is 3 meters from the ground, Mike throws the ball at a height of 2 meters. What velocity does he need in order to hit the backboard 20 cm above the hoop.
distance horizontal=4.5m
vertical path:
hf=hi+vi*sinTheta*t-4.9t^2
3.20=2+v*sinTheta*t-4.9t^2
horizonal path:
4.5=v*cosTheta*t
so you have three unknowns, v, theta, and t. There is not a unique solution.
To determine the velocity needed for Mike to hit the backboard 20 cm above the hoop, we can use the principles of projectile motion.
First, let's define the variables:
- d: Distance from the free throw line to the backboard (4.5 m)
- h: Height of the backboard above the hoop (20 cm or 0.2 m)
- g: Acceleration due to gravity (9.8 m/s^2)
- y: Initial height of the ball when thrown (2 m)
- x: Horizontal distance traveled by the ball
- θ: Angle of launch
The key idea is that the vertical position of the ball when it hits the backboard should be 20 cm above the height of the hoop. This means the overall height the ball needs to travel is 3 m (hoop height) + 0.2 m (backboard height) = 3.2 m.
We can start by calculating the time it takes for the ball to reach its peak height. At that point, the vertical velocity component will be zero. Using the kinematic equation:
v_y = u_y + gt
Since the final vertical velocity (v_y) is zero at the top of the trajectory, and the initial vertical velocity (u_y) is determined by the launch angle (θ) and initial velocity (v), we can solve for the time (t) it takes to reach the peak height.
0 = v * sin(θ) - g * t
Solving for t:
t = (v * sin(θ)) / g
Next, we can determine the time it takes for the ball to travel from the peak height to the backboard. This will be the same as the time it takes to travel from the backboard to the hoop, as they are symmetrical paths.
Using the kinematic equation for the vertical distance traveled:
s_y = u_y * t + (1/2) * g * t^2
Substituting the known values:
0.2 m = (v * sin(θ)) * t - (1/2) * g * t^2
Simplifying the equation:
0.2 = (v * sin(θ)) * ([v * sin(θ)] / g) - (1/2) * g * ([v * sin(θ)] / g)^2
0.2 = ((v * sin(θ))^2) / g - ((v * sin(θ))^2) / (2g)
Multiplying both sides by 2g to eliminate the denominator:
0.4g = (v * sin(θ))^2 - (v * sin(θ))^2/2
0.4g = (v * sin(θ))^2 * (1 - 0.5)
0.4g = (v * sin(θ))^2 * 0.5
(v * sin(θ))^2 = (0.4g) / 0.5
(v * sin(θ))^2 = 0.8g
Taking the square root of both sides:
v * sin(θ) = √(0.8g)
Solving for v:
v = √(0.8g) / sin(θ)
Now, we can plug in the known values to find the required velocity:
v = √(0.8 * 9.8) / sin(θ)
v = √7.84 / sin(θ)
v ≈ 2.8 / sin(θ)
Please note that we still need to know the angle of launch (θ) to find the exact velocity needed for Mike to hit the backboard 20 cm above the hoop.