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 = 9.00 m from the basket, as in Figure P4.54. If he shoots the ball at a 33.0° 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.

Explanation / Answer

let V be the intial speed of the ball

There are two speed component. One is invertical speed and one is horizonal speed

let Vx = horizonal speed
Vy = vertical speed.

Vx = Vcos33.0°
Vy = Vsin33.0°

The displacement of the ball in horizonal direction
x = vt
9= Vcos33.0° t

the time it takes the ball to go throught the basket vertically is also the time it takes the ball to go through the ball horizonally.

solve for t
t = 9 / Vcos33.0°

the distance placement of the ball in vertical direction
Xf = .5at^2 + Vt + Xi

we know that t = 9 / Vcos33.0°
3.05 = .5(-9.8) (9 / Vcos33.0°)^2 + Vsin33.0°(9 / Vcos33.0°) + 2

now just solve:
1.05 = .5(-9.8) (9 / Vcos33.0°)^2 + Vsin33.0° (9 / Vcos33.0°)
1.05 = -4.9 (9 / Vcos33.0°)^2 + 5.844
-4.794= -4.9 (9 / Vcos33.0°)^2
0.978 = (9 / Vcos33.0°)^2
0.988 = 9 / Vcos33.0°
0.988 Vcos33.0° = 9
V = 10 / (0.988 cos33.0°)
V = 12.06 m/s

so the intial velocity of the ball is about 10.67 m/s

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