A basketball player attempts a three-point shot 10.0 m from the basket as shown

Question

A basketball player attempts a three-point shot 10.0 m from the basket as shown in the figure below. The player shoots the ball at an angle of = 39.29 from horizontal, and releases the ball at a height of h 1.84 m. The rim of the basket is at a height of H 3.05 m 10.0 m (a) What is the acceleration (in m/s2) of the basketball at the highest point in its trajectory? magnitude 9.8 m/s2 direction downward (b) At what speed (in m/s) must the player throw the basketball so that the ball goes through the hoop without striking the backboard? 1.49 How can you relate the time of travel to the horizontal distance and initial speed? To the vertical displacement and initial speed? m/s

Explanation / Answer

b)

consider the motion along the X-direction

Vox = initial velocity = vo Cos39.2

X = horizontal displacement = 10

t = time of travel = ?

using the equation

X = Vox t

10 = (vo Cos39.2) t

t = 10 /(vo Cos39.2) eq-1

consider the motion along the Y-direction

Yo = initial position = h = 1.84 m

Voy = initial velocity = vo Sin39.2

a = acceleration = - 9.8

Y = final position = H = 3.05 m  

t = time of travel = 10 /(vo Cos39.2)

using the equation

Y = Yo + Voy t + (0.5) a t2

3.05 = 1.84 + vo Sin39.2 (10 /(vo Cos39.2) ) + (0.5) (- 9.8) (10 /(vo Cos39.2) )2

3.05 = 1.84 + Sin39.2 (10 /(Cos39.2) ) + (0.5) (- 9.8) (10 /(vo Cos39.2))2

vo = 26.6 m/s

t = 2 vo Sin /g

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