Question

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

Answer 1

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)?

Answer:

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.

Answer 2

Complete Question

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

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

Answer:

a The maximum height is  $s =2.64m$

b The mass would not effect the height because  by making s the subject in the equation of motion    

                    $s = \frac{v^2 -u^2}{2g}$

We see that it is not dependent on mass

Explanation:

From the question we are told that

        The mass of the baseball is $m_b = 0.6kg$

       The speed it leave hands is $v_h = 9m/s$

       The speed it slow down to is $v_s = 5.4m/s$

According to the law of motion

                  $v_s^2 = v_h^2 + 2as$

Now a is  the acceleration and in this question is the acceleration due to gravity denoted as $-g$ the negative sign shows that it is moving against gravitational force

        Substituting this into the equation

                    $v_s^2 = v_h^2 -2gs$

Substituting values and solving for s we have

                 $5.4^2 = 9^2 -(2* 9.8 *s)$

                 $19.6s = 51.84$

                       $s = \frac{51.84}{19.6}$

                         $s =2.64m$