Calculating Initial Velocity for Accurate Basketball Shot

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Alserina
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Homework Statement


A basketball leaves a player's hands at a height of 2.15m above the floor. The basket is 3.05m above the floor. The player likes to shoot the ball at a 40-degree angle.

If the shot is made from a horizontal distance of 10.00m and must be accurate to +0.32m (horizontally), what is the range of initial speeds allowed to make the basket?


Homework Equations


sinx = opposite side/hypotenuse


The Attempt at a Solution


What I attempted to do was draw two right-angled triangles /| with a 40-degree angle, a 0.9m vertical side and a variable horizontal side (9.68m/10.32m) and tried to figure out the hypotenuse, but it doesn't seem right...
 
on Phys.org
final velocity = initial velocity + (9.80)(time)? I don't really get which p.m. equation would apply...
 
Alserina said:
final velocity = initial velocity + (9.80)(time)? I don't really get which p.m. equation would apply...
The equation can be written as
y = yo + x*tanθ - 1/2*g*x2/(v*cosθ)^2
Solve for v.