A ball is thrown vertically upward from the ground with a speed of 36.0 m/s.Suppose the acceleration of gravity were only 2.00 m/s^2 instead of 9.80 m/s^2

a)At what time does it have a velocity of 12.0 m/s upward?
b)At what time does it have a velocity of 12.0 m/s downward?
c)When is the velocity of the ball zero?
d)When is the displacement of the ball zero?
e)What are the magnitude and direction of the acceleration while the ball is moving upward?
f)What are the magnitude and direction of the acceleration while the ball is moving downward?
g)What are the magnitude and direction of the acceleration when it is at highest point?

v(t)=vi-9.8t

d(t)=+vi*t-4.9t^2

To solve these problems, we need to use the equations of motion for uniformly accelerated motion. The equations of motion we will use are:

- v = u + at
- s = ut + (1/2)at^2
- v^2 = u^2 + 2as

Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time taken
- s is the displacement

Given:
u = 36.0 m/s
a = 2.00 m/s^2

a) At what time does it have a velocity of 12.0 m/s upward?

We can rearrange the equation v = u + at to solve for time:
t = (v - u) / a

Substituting the given values:
t = (12.0 m/s - 36.0 m/s) / 2.00 m/s^2
t = -12.0 s / 2.00 m/s^2
t = -6.00 s

The negative sign indicates that the ball is moving upward, against the gravitational acceleration.

b) At what time does it have a velocity of 12.0 m/s downward?

Using the same equation as above:
t = (v - u) / a

Substituting the given values:
t = (12.0 m/s - 36.0 m/s) / 2.00 m/s^2
t = -24.0 s / 2.00 m/s^2
t = -12.0 s

Again, the negative sign indicates that the ball is moving downward, in the same direction as the gravitational acceleration.

c) When is the velocity of the ball zero?

To find the time when the velocity is zero, we can equate the equation v = u + at to zero:
0 = 36.0 m/s + 2.00 m/s^2 * t

Solving for t:
t = -36.0 m/s / 2.00 m/s^2
t = -18.0 s

Again, the negative sign indicates that the ball is moving upward.

d) When is the displacement of the ball zero?

The displacement will be zero when the ball reaches its highest point and starts coming back down. Therefore, we need to find the time it takes to reach the maximum height, and then double that time.
Let's find the time it takes to reach the maximum height:

Using the equation v = u + at, and substituting in the known values:
0 = 36.0 m/s - 2.00 m/s^2 * t

Solving for t:
t = 36.0 m/s / 2.00 m/s^2
t = 18.0 s

This is the time taken to reach the maximum height. To find the total time for the displacement to be zero, we double this value:
t = 2 * 18.0 s
t = 36.0 s

Therefore, the displacement of the ball is zero at 36.0 seconds.

e) What are the magnitude and direction of the acceleration while the ball is moving upward?

The magnitude of the acceleration while the ball is moving upward is 2.00 m/s^2, as given in the problem.

The direction of the acceleration is opposite to the direction of motion, so it is downward.

f) What are the magnitude and direction of the acceleration while the ball is moving downward?

The magnitude of the acceleration while the ball is moving downward is also 2.00 m/s^2, as the acceleration due to gravity is constant.

The direction of the acceleration is again downward, since it is in the same direction as the gravitational acceleration.

g) What are the magnitude and direction of the acceleration when it is at the highest point?

At the highest point, the velocity of the ball is zero. Therefore, there is no acceleration acting on the ball at the highest point.

The magnitude of the acceleration is 0 m/s^2, and the direction is undefined or nonexistent.

Keep in mind that in reality, the acceleration due to gravity is approximately 9.80 m/s^2, not 2.00 m/s^2 as given in this problem.

To solve these problems, we can use the equations of motion and the laws of physics. First, let's define the given information:

Initial velocity (u) = 36.0 m/s (upward)
Gravity acceleration (g) = 2.00 m/s^2

a) To find the time (t) when the velocity is 12.0 m/s upward:
We can use the equation of motion: v = u + at
Rearranging the equation, we have t = (v - u) / a
Substituting the given values, we get:
t = (12.0 m/s - 36.0 m/s) / (-2.00 m/s^2)
t = (-24.0 m/s) / (-2.00 m/s^2)
t = 12.0 s

Therefore, the ball has a velocity of 12.0 m/s upward at 12.0 seconds.

b) To find the time (t) when the velocity is 12.0 m/s downward:
Using the same equation as above, but this time with negative values for the velocity and acceleration, we get:
t = (12.0 m/s - 36.0 m/s) / (2.00 m/s^2)
t = (-24.0 m/s) / (2.00 m/s^2)
t = -12.0 s

However, time cannot be negative in this case, as it represents a physical quantity. Therefore, there is no valid solution for this question.

c) To find the time (t) when the velocity of the ball is zero, we use the equation v = u + at:
0 = 36.0 m/s - 2.00 m/s^2 * t
Solving for t, we get:
t = 36.0 m/s / 2.00 m/s^2
t = 18.0 s

Therefore, the velocity of the ball is zero at 18.0 seconds.

d) Displacement (s) is the position of an object relative to its starting point. To find the time when the displacement is zero, we need to find the highest point of the ball's trajectory. At the highest point, the velocity is zero. Therefore, the displacement is zero when the velocity is zero.
Referring back to part c), the displacement is zero at 18.0 seconds.

e) The acceleration of a freely falling object near the Earth's surface is always directed downward and is equal to the acceleration due to gravity, which is 2.00 m/s^2 in this case.
Therefore, the magnitude of acceleration when the ball is moving upward is 2.00 m/s^2, directed downward.

f) Similarly, when the ball is moving downward, the magnitude of the acceleration remains 2.00 m/s^2, directed downward.

g) At the highest point of the ball's trajectory, the velocity is zero. Therefore, the acceleration is still acting downward with a magnitude of 2.00 m/s^2 and directed downward.