A Boston Celtics basketball player who is 2.00 m tall is standing on the floor L

Question

A Boston Celtics basketball player who is 2.00 m tall is standing on the floor L = 11.0 m from the basket, as in Figure P4.54. If he shoots the ball at a 39.0° angle with the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m. Hints: The ending height is not the same as the starting height, so you cannot use the range or time of flight formulas. Decompose the motion into x and y components. Using the x-component of the motion, write an expression that relates v0 and t. Then do the same for the y-component of motion. You should then have two equations for the two unknowns v0 and t.
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Explanation / Answer

along x direction

vox = vo*cotheta

acceleration ax = 0

displacement = x

x = vox*t

t = x/vox ,......(1)

along y direction

initial velocity = voy = vo*sintheta

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

initial position = yo = 2m

final position = y = 3.05

dispalcement = y - yo = voy*t + 0.5*g*t^2

y-yo = voy*x/vox - 0.5*g*x^2/(vo^2*(costheta)^2)

y-yo = x*tantheta - 0.5*g*x^2/(vo^2*(costheta)^2)

given theta = 39

x = L = 11

3.05 - 2 = 11*tan39 - (0.5*9.8*11^2/(vo^2*(cos39)^2))

vo = 11.2 m/s   <<<----answer

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