A 0.60kg basketball is dunked vertically downward so that it leaves the [layer's hand at 8m/s. Assume it is released when the bottom of the ball is at the rim (3.05 meters off the ground).
a). find acceleration of the ball once it leaves the player's hand
b). how ling will it take the ball to reach the ground.
c). what will be the velocity of the ball when it hit the ground.
To solve this problem, we can use the equations of motion under constant acceleration. Let's break down each part of the problem step by step.
a) Finding the acceleration of the ball once it leaves the player's hand:
We can start by identifying the initial velocity (u), final velocity (v), displacement (s), and acceleration (a).
Given:
Mass of the basketball (m) = 0.60 kg
Initial velocity (u) = 8 m/s
Displacement (s) = 3.05 m (height of the rim)
The final velocity when the ball reaches the rim is zero since it stops momentarily before falling. Therefore:
Final velocity (v) = 0 m/s
We can use the equation of motion: v^2 = u^2 + 2as
Rearranging the equation, we get:
a = (v^2 - u^2) / 2s
Plugging in the values:
a = (0^2 - 8^2) / (2 * 3.05)
Simplifying the equation, we have:
a = (-64) / 6.1
a ≈ -10.49 m/s²
The acceleration of the ball once it leaves the player's hand is approximately -10.49 m/s². The negative sign indicates that the acceleration is directed downwards.
b) Finding how long it will take the ball to reach the ground:
To find the time (t) it takes for the ball to reach the ground, we can use the equation of motion: s = ut + (1/2)at^2
Since the ball starts from rest at the rim, the initial velocity (u) is zero. Therefore, the equation simplifies to:
s = (1/2)at^2
Plugging in the values:
3.05 = (1/2) * (-10.49) * t^2
Multiplying both sides by 2 and dividing by -10.49, we get:
t^2 ≈ -3.05 * 2 / 10.49
t^2 ≈ -0.5812
The time cannot be negative, so there must be an error. It seems that we forgot to consider the negative sign of acceleration in the equation. We need to take the absolute value of the acceleration to find the correct answer.
Repeating the calculation:
t^2 ≈ 3.05 * 2 / 10.49
t^2 ≈ 0.5812
Now we can find the square root to get the time:
t ≈ √0.5812
t ≈ 0.762 s (approximately)
Therefore, it will take approximately 0.762 seconds for the ball to reach the ground.
c) Finding the velocity of the ball when it hits the ground:
We can use the kinematic equation: v = u + at to find the final velocity (v) of the ball when it hits the ground.
Plugging in the values:
v = 0 + (-10.49) * 0.762
v ≈ -7.99 m/s
The negative sign indicates that the velocity is directed downwards.
Therefore, the velocity of the ball when it hits the ground is approximately -7.99 m/s.