A player passes a 0.600- basketball downcourt for a fast break. The ball leaves the player's hands with a speed of 8.10 and slows down to 7.20 at its highest point.Ignoring air resistance, how high above the release point is the ball when it is at its maximum height?
To find the height above the release point when the ball is at its maximum height, we can use the principles of projectile motion.
The first step is to determine the time it takes for the ball to reach its highest point. We can do this by finding the time it takes for the ball to reach its maximum height and then double it. This is because the time taken to reach the highest point is equal to the time taken to reach the highest point from the maximum height back to the release point.
Given:
Initial velocity (u) = 8.10 m/s
Final velocity (v) = 7.20 m/s
The change in velocity, Δv, is given by:
Δv = v - u
= 7.20 - 8.10
= -0.90 m/s (negative sign indicates a decrease in velocity)
Using the equation for vertical motion under constant acceleration:
Δv = g × t
where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time taken to reach the maximum height.
-0.90 = -9.8 × t
Solving for t:
t = -0.90 / -9.8
t ≈ 0.092 seconds
To find the height above the release point, we can use the equation for vertical displacement:
s = u × t + (1/2) × g × t²
Substituting the known values:
s = 8.10 × 0.092 + 0.5 × 9.8 × 0.092²
s ≈ 0.734 meters
Therefore, the ball is approximately 0.734 meters above the release point when it is at its maximum height.