a ball moves up a hill with an initial velocity of 3m/s. four seconds later, it is moving down the hill at 9.0m/s. find the displacement of the ball from its initial point of release

see other post.

To find the displacement of the ball, we need to determine the distance between its initial position and its final position.

Given:
Initial velocity, u = 3 m/s (uphill)
Velocity after 4 seconds, v = -9 m/s (downhill)

Since the ball changes direction, we know that its final velocity is the negative of its initial velocity.

First, let's determine the acceleration of the ball. We can use the equation:

v = u + at

Rearranging this equation, we have:

a = (v - u) / t

Substituting the values, we get:

a = (-9 m/s - 3 m/s) / 4 s
a = -12 m/s / 4 s
a = -3 m/s²

Now, let's find the time it takes for the ball to change direction. To do this, we can use the equation:

v = u + at

Rearranging this equation, we have:

t = (v - u) / a

Substituting the values, we get:

t = (-9 m/s - 3 m/s) / (-3 m/s²)
t = -12 m/s / -3 m/s²
t = 4 s

Now that we have the time it takes for the ball to change direction, we can find its displacement using the equation:

s = ut + (1/2)at²

Since the initial velocity is uphill, the displacement can be calculated as:

s = 3 m/s * 4 s + (1/2) * (-3 m/s²) * (4 s)²
s = 12 m + (-3/2) m/s² * 16 s²
s = 12 m + (-48/2) m²/s²
s = 12 m - 24 m²/s²
s = -12 m

Therefore, the displacement of the ball from its initial point of release is -12 m.