Ill do the rest.

A ball was thrown up at 25.0 m/s. It was released when it was 2.0 m above the ground.
a) For how long did the ball rise?
b) at the top of its trajectory, how far above the ground was it?
c) what was the ball's speed, direction and height 2.0 s after release?
d) what was its speed, direction and height after 4.0 s after release?

v = Vi - 9.8 t

when is v zero (stopped at top) ?
0 = 25 - 9.8 t
solve for t

Thank you very much. :)

You are welcome :)

To solve these questions, we can utilize the equations of motion. Let's go step by step:

a) For how long did the ball rise?
To find the time it takes for the ball to reach its maximum height, we can use the equation:

v = u + at

Where:
v = final velocity (0 m/s when the ball reaches its maximum height)
u = initial velocity (25.0 m/s)
a = acceleration (in this case, acceleration due to gravity, which is approximately 9.8 m/s²)
t = time (unknown)

Rearranging the equation, we get:

t = (v - u) / a

Substituting the values, we have:

t = (0 - 25.0) / (-9.8)
t = -25.0 / -9.8
t ≈ 2.55 seconds

Therefore, the ball takes approximately 2.55 seconds to reach its maximum height.

b) At the top of its trajectory, how far above the ground was it?
To find the height at the highest point, we can use the equation:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s)
u = initial velocity (25.0 m/s)
a = acceleration (in this case, acceleration due to gravity, which is approximately -9.8 m/s² since it acts in the opposite direction)
s = displacement (unknown)

Rearranging the equation, we get:

s = (v^2 - u^2) / (2a)

Substituting the values, we have:

s = (0 - (25.0)^2) / (2 * -9.8)
s = -625.0 / -19.6
s ≈ 31.88 meters

Therefore, at the top of its trajectory, the ball is approximately 31.88 meters above the ground.

c) What was the ball's speed, direction, and height 2.0 seconds after release?
To find the ball's speed, direction, and height after a given time, we need to consider its motion both upward and downward.

Speed:
The speed at any given time can be found using the equation:

v = u + at

For the upward motion (t = 2.0 seconds):

v = 25.0 - 9.8 * 2.0
v = 25.0 - 19.6
v ≈ 5.4 m/s

For the downward motion (t = 2.0 seconds):

v = u + at
v = 0 + 9.8 * 2.0
v = 0 + 19.6
v ≈ 19.6 m/s (since it is in the opposite direction)

Therefore, the speed of the ball after 2.0 seconds is approximately 5.4 m/s upward and 19.6 m/s downward (negative sign indicates direction).

Height:
The height at any given time can be found using the equation:

s = ut + (1/2)at^2

For the upward motion (t = 2.0 seconds):

s = 25.0 * 2.0 + (1/2) * (-9.8) * (2.0)^2
s = 50.0 - 19.6
s ≈ 30.4 meters

For the downward motion (t = 2.0 seconds):

s = 0 + 19.6 * 2.0 + (1/2) * (-9.8) * (2.0)^2
s = 39.2 - 19.6
s ≈ 19.6 meters (since it is in the opposite direction)

Therefore, the height of the ball after 2.0 seconds is approximately 30.4 meters upward and 19.6 meters downward (negative sign indicates direction).

d) What was its speed, direction, and height after 4.0 seconds after release?
Similarly, to find the speed, direction, and height after 4.0 seconds, we can use the same equations mentioned above.

Speed:
For the upward motion (t = 4.0 seconds):

v = 25.0 - 9.8 * 4.0
v = 25.0 - 39.2
v ≈ -14.2 m/s (negative sign indicates direction)

For the downward motion (t = 4.0 seconds):

v = 0 + 9.8 * 4.0
v = 0 + 39.2
v ≈ 39.2 m/s

Therefore, the speed of the ball after 4.0 seconds is approximately 14.2 m/s upward and 39.2 m/s downward.

Height:
For the upward motion (t = 4.0 seconds):

s = 25.0 * 4.0 + (1/2) * (-9.8) * (4.0)^2
s = 100.0 - 78.4
s ≈ 21.6 meters

For the downward motion (t = 4.0 seconds):

s = 0 + 39.2 * 4.0 + (1/2) * (-9.8) * (4.0)^2
s = 156.8 - 78.4
s ≈ 78.4 meters

Therefore, the height of the ball after 4.0 seconds is approximately 21.6 meters upward and 78.4 meters downward.