Problem 2 A 6 ft tall person is standing 632 m away a 10 ft tall basketball hoop

Question

Problem 2 A 6 ft tall person is standing 632 m away a 10 ft tall basketball hoop. If launch the ball at 50 degree with respect to horizontal from just above their head towards the hoop, what is the initial velocity they must give the ball in order to make the shot. (Ignore air resistance.) (1 ft = 0.305 m) Answer: 13.64 m/s 11.58 m/s 8.65 m/s 7.65 m/s 5.23 m/s Follow up: Now consider the same case but with air resistance present and proportional to the velocity squared. What happens the to direction of the final velocity compared to the case with no air resistance? (You do not need to solve exact values.)

Explanation / Answer

given,

height of person = 6ft or 1.83 m

height of basketball hoop = 10 ft or 3.05 m

angle of launch = 50 degree

so,

speed = distance / time

horizontal velocity = v * cos(50)

v * cos(50) = 6.32 / t            (1)

vertical velocity = v * sin (50)

s = v * sin (50) * t - 0.5 * g * t^2

from 1 and 2 we get

(3.05 - 1.83) = v * sin(50) * 6.32 / v * cos(50) - 0.5 * 9.8 * (6.32 / v * cos(50))^2

velocity = 8.65 m/s

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