A player passes a 0.600- basketball downcourt for a fast break. The ball leaves the player's hands with a speed of 8.40 and slows down to 7.20 at its highest point.

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To find the time it takes for the basketball to reach its highest point, we can use the equation of motion:

v² = u² + 2ax

Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
x is the displacement.

At the highest point, the velocity is zero (v = 0), the initial velocity is 8.40 m/s (u = 8.40 m/s), and the acceleration due to gravity is acting downward (-9.8 m/s²). We can now solve for the displacement:

0 = (8.40)² + 2(-9.8)x

Simplifying the equation:

0 = 70.56 - 19.6x

Rearranging the equation:

19.6x = 70.56

x = 70.56 / 19.6
x ≈ 3.60 m

Therefore, the basketball reaches a height of approximately 3.60 meters.

To find the time it takes for the basketball to reach its highest point, we can use the equation of motion:

v = u + at

Where:
v is the final velocity (0 m/s),
u is the initial velocity (8.40 m/s),
a is the acceleration (-9.8 m/s²), and
t is the time.

Rearranging the equation:

0 = 8.40 + (-9.8)t

-8.40 = -9.8t

Dividing both sides by -9.8:

t ≈ 8.40 / 9.8
t ≈ 0.857 seconds

Therefore, it takes approximately 0.857 seconds for the basketball to reach its highest point.

To find the time it takes for the basketball to slow down from 8.40 m/s to 7.20 m/s, we can use the equation of motion:

v = u + at

Where:
v is the final velocity (7.20 m/s),
u is the initial velocity (8.40 m/s),
a is the acceleration, and
t is the time.

We need to find the acceleration first. Rearranging the equation:

v = u + at
7.20 = 8.40 + a*t

Subtracting 8.40 from both sides:

-1.20 = a*t

Now we need to find the acceleration (a). We can use the equation:

a = (v - u) / t

Substituting the given values:

a = (7.20 - 8.40) / t
a = -1.20 / t

Since the acceleration due to gravity (-9.8 m/s²) is acting upward, the acceleration (a) is equal to -9.8 m/s². Therefore:

-9.8 = -1.20 / t

To solve for t, we can rearrange the equation:

t = -1.20 / -9.8
t ≈ 0.122 seconds

Therefore, it takes approximately 0.122 seconds for the basketball to slow down from 8.40 m/s to 7.20 m/s.

Note: If you need any clarification or further assistance, feel free to ask!

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