A basketball leaves a player's hands at a height of 2.10 m above the floor. The basket is 3.05 mabove the floor. The player likes to shoot the ball at a 40.0 ∘ angle.
If the shot is made from a horizontal distance of 6.60 m and must be accurate to ±0.23m (horizontally), what is the range of initial speeds allowed to make the basket?
Case 1 :
X = horizontal distance
vox = component of launch velocity along x-direction = vo Cos40
t = time of travel
Time of travel is given as
t = X/vox (Since there is no acceleration along x-direction)
t = X/(vo Cos40)
Consider the motion along the Y-direction :
Yo = initial position = 2.10 m
Yf = final position = 3.05 m
a = acceleration = - 9.8 m/s2
voy = component of launch velocity along x-direction = vo Sin40
t = time of travel = X/(vo Cos40)
using the equation
Yf = Yo + voy t + (0.5) a t2
3.05 = 2.10 + (vo Sin40) (X/(vo Cos40)) + (0.5) (- 9.8) (X/(vo Cos40))2
3.05 = 2.10 + (Sin40) (X/(Cos40)) + (0.5) (- 9.8) (X/(vo Cos40))2
when X = 6.60 - 0.23 = 6.37 m
3.05 = 2.10 + (Sin40) (6.37/(Cos40)) + (0.5) (- 9.8) (6.37/(vo Cos40))2
vo = 8.8 m/s
when X = 6.60 + 0.23 = 6.83 m
3.05 = 2.10 + (Sin40) (6.83/(Cos40)) + (0.5) (- 9.8) (6.37/(vo Cos40))2
vo = 8.42 m/s
so the range is from 8.42 m/s to 8.8 m/s
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