# A guy 1.97 m tall is 11.6 m away from a hoop at a height of 3.05 m.If he shoots the ball at a 40.2degre angle, what is his intial velocity so that he scores?

If the ball is released with velocity v at (x, y) = (0, 1.97) and must pass through the point (11.6, 3.05), the horizontal and vertical equations of motion can be written as
x = v*because(40.2°)*t
y = 1.97 + v*sin(40.2°)*t - (1/2)*9.8*t^2
where the acceleration due to gravity is taken to be -9.8 m/s^2.

We can solve the first equation for t and substitute that expression into the second equation. At the hoop position, we have
11.6 = v*.763796*t
15.1873/v = t    (divide by the coefficient of t)

3.05 = 1.97 + v*.645458*(15.1873/v) - 4.9*(15.1873/v)^2
3.05 = 1.97 + 9.80276 - 1130.205/v^2    (simplify)
-8.72276 = -1130.205/v^2    (subtract the constant terms)
v^2 = 1130.205/8.72276    (multiply by v^2/-8.72276)
v = √129.570    (take the square root)
v = 11.38 m/s _____
The ball is in the air about 15.1873/11.3829 = 1.334 seconds, computed using our expression for t.
thanked the writer.