Player passes a 0.6 kg basketball downcourt for a fast break. The ball leaves the player's hands with a speed of 9 m/s and slows down to 5.4 m/s at its highest point. a.) How high (above the release point) is the ball when it is at its maximum height

The second part of the question is missing and it says;

b) How would doubling the ball's mass affect the result in part (a)?

A) Height = 2.645m

B) Doubling the mass wouldn't affect the result in part a.

Explanation:

A) Initial speed u = 9 m/s

Final speed v = 5.4 m/s

Now, kinetic energy is given as;

Initial kinetic energy; K1 = (1/2)mu²

Final kinetic energy; K2 = (1/2)mv²

Change in kinetic energy = K2 - K1 = (1/2)mv² - (1/2)mu² = (1/2)m(v² - u²)

Let the ball move to a height h.

Thus, Change in potential energy = mg(h-0) = mgh

conservation of energy is given as; change in kinetic energy + change in potential energy = 0

Thus,

(1/2)m(v² - u²) + mgh = 0

Dividing through by m, to get;

(1/2)(v² - u²) + gh = 0

Multiplying through by 2,

v² - u² + 2gh = 0

Thus: v² - u² = -2gh

u² - v² = 2gh

h = (u² - v²)/(2g)

Substitute values: -

h = (9² - 5.4²)/(2x9.8)

= 51.84/19.6

= 2.645m

B) From the steps above, we can see that the mass wasn't used to calculate the height. Thus, even if it is doubled, it doesn't affect the height.

c

explanation:

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