Bill throws a tennis ball vertically upward such that it leaves his hand with a speed of 12 m/s. Bill catches the ball at the same height from which he released it. Assume that air resistance is negligible, so the tennis ball can be considered to be in free-fall, with an acceleration of magnitude g=9.80 m/s^2. Determine the maximum vertical displacement of the ball and the time it takes to reach that point.

2.25 i dun know

To determine the maximum vertical displacement of the ball and the time it takes to reach that point, we can use the equations of motion for vertical motion in free-fall.

Let's start by finding the time it takes for the ball to reach its maximum height. We know that the initial velocity of the ball (u) is 12 m/s, and the acceleration due to gravity (g) is -9.80 m/s^2 (negative because it's directed downwards). The final velocity at maximum height (v) will be 0 m/s because the ball momentarily stops before reversing direction.

We can use the equation v = u + gt, where:
v = 0 m/s (final velocity)
u = 12 m/s (initial velocity)
g = -9.80 m/s^2 (acceleration due to gravity)
t = time taken

Rearranging the equation to solve for t, we have:
t = (v - u) / g

Substituting the given values:
t = (0 - 12) / -9.80
t = 12 / 9.80
t ≈ 1.22 seconds

So, it takes approximately 1.22 seconds for the ball to reach its maximum height.

Next, let's determine the maximum height (vertical displacement) of the ball. We can use the equation s = ut + 0.5gt^2, where:
s = vertical displacement (maximum height)
u = 12 m/s (initial velocity)
t = 1.22 seconds (time taken)
g = -9.80 m/s^2 (acceleration due to gravity)

Substituting the given values:
s = (12)(1.22) + 0.5(-9.80)(1.22)^2
s ≈ 7.41 meters

Therefore, the maximum vertical displacement of the ball is approximately 7.41 meters.