Question

I need a plan to solve this problem and I'm stuck. Please help.
A basketball laeaves a player's hands at a height of 2.10m above the floor. The basket is 2.60m above the floor. The player shoots the ball at a 38.5 degree angle. If the shot is made from a horizontal distance of 12.0m and must be accurate to +or- 0.22 (horizontally), what is the range of initial speeds allowed to make the basket?

Answer 1

Derive the formula for horizontal "range" distance R travelled between the two heights, as a function of V. Then find the values of V that correspond to 12.22 and 11.78 m range for R.

R = V cos 38.5 * T , where T is the time of flight between the two heights.

Assume the ball must be coming down, not going up, when it reaches 2.6 m height.

Y = 2.1 + V sin 38.5 T - 4.9 T^2 = 2.6
4.9 T^2 - 0.623 VT + 0.5 = 0
T = [0.623 + sqrt(0.388V^2 - 9.8)]/9.8
I have not used the smaller of the two roots of the quadratic equation because that corresponds to the ball going UP.

The horizontal distance travelled in time T is
R = V cos 38.5 T = 0.783 VT
R = 0.0799V[0.623 + sqrt(0.388V^2- 9.8)]

Check my math and thinking. You may have to use graphical or iterative means to find the value of V that correspond to R = 12.22 and 11.78 m