A ball thrown vertically upward is caught by the thrower after 2 seconds. Find the initial velocity of the ball and the maximum height the ball reaches???!!

so it took 1 second to get to the top.

Consider it coming down...

vf=-gt so it looks like at one second it is going 9.8m/s
h=1/2 g t^2

To find the initial velocity (u) of the ball, as well as the maximum height (h) it reaches, we can use the equations of motion for vertical motion.

Let's break down the problem:

1. The ball is thrown vertically upward and caught after 2 seconds.
2. We need to determine the initial velocity (u) and the maximum height (h).

To find the initial velocity of the ball (u), we can use the following equation:

v = u + gt

Where:
- v is the final velocity (which is 0 when the ball reaches its maximum height).
- g is the acceleration due to gravity (approximately 9.8 m/s²).
- t is the time taken (which is 2 seconds).

Substituting the given values into the equation, we have:

0 = u + (9.8 * 2)

Simplifying the equation, we get:

0 = u + 19.6

Rearranging the equation, we find:

u = -19.6 m/s

The negative sign indicates that the initial velocity is in the opposite direction of the acceleration due to gravity, which makes sense since the ball is thrown upward.

Now, let's find the maximum height (h) the ball reaches.

To find the maximum height, we use the following equation:

h = ut + (1/2)gt²

Substituting the values we know:

h = (-19.6 * 2) + (0.5 * 9.8 * 2²)

Simplifying the equation, we have:

h = -39.2 + 19.6

h = -19.6

Ignoring the negative sign, we find that the maximum height is 19.6 meters.

So, the initial velocity of the ball is -19.6 m/s (upwards) and the maximum height it reaches is 19.6 meters.