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So, plus or minus A LOT of error: 

vy0 = 0 

\delta y = 200 ft

\delta x = 60ft

CONVERT TO METRIC 

g = - 9.8 m/s 

t = (vyo - sqrt(vyo^2 - 2*g* \delta y))/g = 3.5s 

vxo = vxf = \delta x/ t = 5.18 m/s 

vyf = vyo - gt = 34.57 m/s 

vf = sqrt(vfy^2 + vfx^2) = 34.95 m/s 

p = m*vf = 19.81 N s

If the basket is 17in and the ball is 9.39in in diameter and if he was 60ft away, he had about a .605 degree angle to aim for the basket not accounting for wind, etc.

Don’t trust me on this.

  1. hashtaghashtag posted this