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?
I have gotten to this point but I am unaware of how to calculate the last part:
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 calculate the velocity at which the player should release the ball, you need to solve the equation derived from the equations of motion for vertical motion:
3 = 1.8 + (x sin 50) t - 4.9 t^2
First, let's simplify the equation:
3 = 1.8 + (x sin 50) t - 4.9 t^2
Rearrange the equation to put it in the form of a quadratic equation:
4.9 t^2 - (x sin 50) t + (1.8 - 3) = 0
Simplify further:
4.9 t^2 - (x sin 50) t - 1.2 = 0
Now, you can solve this quadratic equation for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 4.9, b = -(x sin 50), and c = -1.2. Plug in these values and calculate t.
Once you have the value of t, you can substitute it back into the equation you derived for the horizontal motion:
4.6 = (x cos 50) t
Plug in the value of t and solve for x.
That will give you the horizontal velocity (x-component) at which the player should release the ball to hit nothing but net.