A ball thrown upward from the a point 7 meters below the bottom of a window rises past the window, which is 2.5 meters high, passing the window in 0.33 seconds. How fast was it going when it reached the bottom of the window, and with what velocity was it initially thrown upward?

To solve this problem, we can use the equations of motion for linear motion with constant acceleration.

Let's denote the initial velocity of the ball as "u" and the acceleration due to gravity as "g" which is approximately 9.8 m/s^2 (assuming no air resistance).

1. Time to reach the top of the window:
The ball travels a distance of 2.5 meters in 0.33 seconds. Since the ball is moving upward, we can use the equation:
s = ut + (1/2)gt^2
2.5 = u(0.33) + (1/2)(-9.8)(0.33)^2
2.5 = 0.33u - 0.54
0.33u = 3.04
u = 9.21 m/s

2. Velocity when reaching the bottom of the window:
To determine the velocity when the ball reaches the bottom of the window, we can use the equation:
v = u + gt
where v is the final velocity (which we need to find), u is the initial velocity (9.21 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken (0.33 seconds from the problem statement).
v = 9.21 + (-9.8)(0.33)
v = 9.21 - 3.24
v = 5.97 m/s

Therefore, the ball was going approximately 5.97 m/s when it reached the bottom of the window, and it was initially thrown upward with a velocity of approximately 9.21 m/s.

To find the velocity when the ball reached the bottom of the window and the initial velocity, we need to use the equations of motion for free-falling objects.

1. To find the velocity when the ball reached the bottom of the window, we can use the equation:
v = u + gt

where:
v = final velocity
u = initial velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken to reach the bottom of the window

Given that the ball passed the window in 0.33 seconds, we can substitute the values:
v = u + (9.8 m/s^2)(0.33 s)

2. To find the initial velocity, we need to determine the height the ball traveled above the window. The total displacement can be calculated as the height of the window plus the distance from the starting point to the bottom of the window:
total displacement = 2.5 m + 7 m

3. We can use the kinematic equation to calculate the time taken for the ball to reach the highest point (when its velocity is zero) above the window:
v = u + gt = 0
solving for t, we get:
t = -u / g

4. The time taken to reach the highest point above the window will be equal to the time taken to reach the bottom of the window. Thus, we can substitute the value into the equation found in step 3:
-u / g = 0.33 s

Solving for the initial velocity (u) in the equation from step 4 will give us the answer.

Please note that the acceleration due to gravity is negative when going upwards, so make sure to consider the sign convention accordingly when plugging in the values.

By following these steps, you should be able to calculate both the velocity when the ball reached the bottom of the window and the initial velocity.