Ignoring ball and hoop diameter, just solve
y(x) = -gsec^2(θ)/2v^2 x^2 + tanθ x + h
when y(4.57) = 3.05
-9.8 sec^2θ/(2*8^2) * 4.57^2 + tanθ (4.57) + 2.48 = 3.05
-1.6 (1+tan^2θ) + 4.57 tanθ - 0.57 = 0
1.6 tan^2θ - 4.57 tanθ + 2.17 = 0
tanθ = .602 or 2.255
θ = 31° or 66°
The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 8.00 m/s, releasing it at a height of 2.48 m above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error.
° above the horizontal
2 answers
Steve,
Do you mind explaining where you derived the y(x) equation?
It looks like something I had before I looked for help on this problem and I am just trying to clarify just in case I see this problem again.
I did use this help to get the problem answered correctly and am just curious now.
Any feedback would be great
Do you mind explaining where you derived the y(x) equation?
It looks like something I had before I looked for help on this problem and I am just trying to clarify just in case I see this problem again.
I did use this help to get the problem answered correctly and am just curious now.
Any feedback would be great