Question

A batter pops a ball straight up (a) If the ball returns to the height from which it was hit 4.5 s later, what was its
initial speed? (b) Does the ball reach its maximum height 2.25s after it was hit, more than 2.25s after it was hit, or
less than 2.25s after it was hit? Explain. (c) Find the maximum height of the ball, as measured from the point
where it was hit.

Answer 1

Answer:

a)  S = V0 t + 1/2 a t^2     distance ball travels in time t

0 = V0 t - 1\2 * 9.8 * t^2       ball returns to original height

V0 = 4.9 * 4.5 = 22.05 m / s      original speed

b) Since V0 = g t   time to decelerate to zero

t = 22.05 / 9.8 =  2.25 sec       time up equals time down

c) S = V0 t + 1/2 a t^2  

S = 22.05 * 2.25 - 9.8 / 2 * 2.25^2 = 24.8     Max height

Check: Avg speed = 22.05 / 2 = 11.025 m/s

Time to fall from max height = 11.025 * 2.25 = 24.8 ft

Answer 2

After a batter pops a ball straight up we have:

a) The initial speed of the ball before it returns to the height from which it was hit 4.5 seconds later, is 22.07 m/s.

b) The ball reaches its maximum height 2.25 s after it was hit.  

c) The maximum height of the ball, s measured from the point where it was hit, is 24.8 m.          

a) The initial speed can be calculated with the following equation:

$v_{f} = v_{0} - gt$

We will use this equation to find the initial speed when the ball is moving upward.

Where:

$v_{0}$: is the initial speed =?

$v_{f}$: is the final speed = 0 (at the maximum height)

g: is the acceleration due to gravity = 9.81 m/s² (it is negative because its direction is in the opposite direction to the movement of the ball when it is going up)

t: is the time when the ball is going up

When the ball is going up, the time is:

$t = \frac{4.5 s}{2} = 2.25 s$

Hence, the initial speed is:

$v_{0} = gt = 9.81 m/s^{2}*2.25 s = 22.07 m/s$

b) The ball reaches its maximum height 2.25 s after it was hit because the ball returns to the height from which it was hit. The rise time is equal to the fall time.  

c) The maximum height of the ball is the following:

$v_{f}^{2} = v_{0}^{2} - 2gh$  

Where:

h: is the maximum height

Then, the maximum height is:

$h = \frac{v_{0}^{2}}{2g} = \frac{(22.07 m/s)^{2}}{2*9.81 m/s^{2}} = 24.8 m$

You can find another example of the vertical launch here: https://brainly.com/question/8982997?referrer=searchResults

I hope it helps you!