I need help understanding this step by step

Question

In slow-pitch softball , the pitcher pitches the ball to the batter in a high arc . Suppose a ball leaves the pitchers hand when it is 2.5 ft above the ground with an upward velocity of 27ft/s. The batter hits the ball when it is 3.5 ft above the ground . How much time elapses between the pitch and the hit ?

Answer 1

Assume the ball was thrown straight up and then fell straight down.

h₀ = height from which the ball leaves the pitcher's hand

v₀ = upward speed at which the ball leaves the pitcher's hand

h = height at which the batter hits the ball

What you need to do is calculate the time it takes for the ball to launch from a height of h₀, go up, then come down to a height of h. In feet, with t in seconds,

h(t) = -16t² + v₀t + h₀

h(t) = -16t² + 27t + 2.5

The time of the throw is t=0. You want to find the time it takes for the ball to reach 3.5 ft. Set h(t)=3.5 and solve for t.

h(t) = 3.5

-16t² + 27t + 2.5 = 3.5

-16t² + 27t - 1 = 0

Solve this last equation for t using the quadratic formula, with

a = -16

b = 27

c = -1

You will get 2 positive solutions for t. The lower solution is the time to reach 3.5 feet on the way up. The higher solution is the time to reach 3.5 feet on the way down. Since the batter hits the ball as it is coming down, the higher solution for t is the answer you need for this problem.

Can you finish from here?

Answer 2

To answer this problem you need to know how long it takes the ball to reach a height of 1 foot (to go from 2.5 ft on the release to 3.5 ft when the batter hits it). You know the upward velocity is 27ft per 1 second. So to find how many seconds it takes to rise 1 foot, you need to divide 27feet by 27 to give a new ratio of 1 foot per a different amount of seconds. To find out how many seconds that is, you must divide your original 1 second by that same 27 to keep your proportion equal.

27ft / 27 = 1ft and 1s / 27 = .037seconds

So between the pitch and the hit, there's an elapsed time of .037 seconds.