A baseball is hit vertically upward and is caught by the catcher 2.68s later. Assume the ball was hit and caught at a height of 1.00m.

what is the velocity of the ball when it is hit off the bat?
what is the balls maximum displacement?

Tr = 2.68/2 = 1.34 s. = Rise time.

V = Vo + g*Tr = 0 @ max. ht.
Vo = -(g*Tr) = -(-9.8*1.34) = 13.13 m/s.

D = h - ho = 1m - 1m = 0

To find the velocity of the ball when it is hit off the bat, we can use the equation for vertical motion:

v = u + gt

Where:
v = final velocity (unknown)
u = initial velocity (unknown)
g = acceleration due to gravity (approximated as 9.8 m/s²)
t = time (2.68 s)

Since the ball is hit vertically upward, its final velocity will be zero when it reaches its maximum height and starts to fall back down. Therefore, the equation becomes:

0 = u - 9.8 * 2.68

We can solve this equation to find the initial velocity (u):

u = 9.8 * 2.68

Calculating this:

u ≈ 26.3 m/s

So, the velocity of the ball when it is hit off the bat is approximately 26.3 m/s.

To find the ball's maximum displacement, we can use the equation for vertical motion:

s = ut + (1/2)gt²

Where:
s = displacement (unknown)
u = initial velocity (26.3 m/s)
g = acceleration due to gravity (approximated as 9.8 m/s²)
t = time (2.68 s)

Since the ball was hit and caught at a height of 1.00 m, the displacement (s) will be the total distance traveled by the ball, which is 2 times the height:

s = 2 * 1.00

Calculating this:

s = 2.00 m

So, the ball's maximum displacement is exactly 2.00 meters.