A) What is the acceleration (in m/s 2 ) of the basketball at the highest point i

Question

A) What is the acceleration (in m/s2) of the basketball at the highest point in its trajectory? ( magnitude & direction)

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?

My Notes Ask Your Teacher 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 0 40.8 from horizon height of H-3.05 m. tal, and releases the ball at a height of h - 2.16 m. The rim of the basket is ata 10.0 m

Explanation / Answer

(a) The acceleration is 9.8 m/s2 at all points on the trajectory.

(b) time for ball to move 10 m, horizontally

t = 10 m / vicos 40.8

The vertical height gainedm, y = 3.05 - 2.16 = 0.89 m

Now, use kinematics equation

y = vyit - 1/2*at2

0.89 = visin 40.8 * 10 m / vicos 40.8 - 1/2 * 9.8 * (10 m / vicos 40.8)2

0.89 = 10*tan 40.8 - 4.9*100 / vi2 cos240.8

0.89 = 8.6317 - 490 / 0.573vi2

-7.7417 = - 490 / 0.573vi2

vi = 10.509 m/s

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