Question

A basketball leaves a player's hands at a height of 2.25 m above the floor. The basket is 3.05 m above the floor. The player likes to shoot the ball at a 36.0 ∘ angle If the shot is made from a horizontal distance of 12.00 m and must be accurate to ±0.32 m (horizontally), what is the range of initial speeds allowed to make the basket?

Enter your answers numerically separated by a comma.

Answer 1

PROJECTILE

along horizontal
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initial velocity vox = v0*costheta

acceleration ax = 0

initial position = xo = 0

final position = x

displacement = x - x0

from equation of motion

x - x0 = v0x*T+ 0.5*ax*T^2

x - x0 = vo*costheta*T

T = (x - x0)/(vo*costheta)......(1)

along vertical
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initial velocity v0y = vo*sintheta

acceleration ay = -g = -9.8 m/s^2

initial position y0 = 0

final position y = 0

from equation of motion

y-y0 = v0y*T + 0.5*ay*T^2 .........(2)

using 1 in 2

y-y0 = (vo*sin(theta)*(x-x0))/(vo*cos(theta)) - (0.5*g*(x-x0)^2)/(vo^2*(cos(theta))^2)

y - y0 = (x - x0)*tantheta - (0.5*g*(x-x0)^2)/(vo^2*(cos(theta))^2)

for y = 3.05

x = 12-0.32 = 11.68 m

theta = 36

yo = 2.25 m

3.05 - 2.25 = (11.68-0)*tan36 - (0.5*9.81*(11.68-0)^2/(v0^2*(cos36)^2))

v0 = 11.53 m/s

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for x = 12 + 0.32 = 12.32 m

3.05 - 2.25 = (12.32-0)*tan36 - (0.5*9.81*(12.32-0)^2/(v0^2*(cos36)^2))

v0 = 11.81 m/s <<<----------answer

11.53 m/s , 11.81 m/s