Isn't this a plug and chug? I will be happy to critique your thinking. Your equal signs in the function are not correct, I assume that is a typo.

suppose a baseball is shot up from thr ground straight up with an initial velosity of 32 feet per second . A function can be created by expressing distance above the ground , s, as a function of time , t. This function is s=16t2 = vot = so. 16 represents 1/2 g , the gravitational pull due to gravity 9 measured in feet per second 2). vo is the initial velosity (how hard do you throw the object , measured in feet per second). so is the initial distance above ground (in feet). if you are standing on the ground , then so=0. what is the function that describes this problem? The ballwill be how high above the ground after 1 second? How long will it take to hit the ground ? what is the maximum height of the ball show the work.

0.08

It seems that there is a misunderstanding in the equation you provided. The equation you wrote, s=16t2 = vot = so, contains some errors and misuse of equal signs. Let me correct it for you:

The correct equation for the distance above the ground, s, as a function of time, t, is:
s = -16t^2 + vot + so

Now, let's answer the questions step-by-step:

1. What is the function that describes this problem?
The function that describes the distance above the ground, s, as a function of time, t, is:
s = -16t^2 + vot + so

2. How high above the ground is the ball after 1 second?
To find the height above the ground after 1 second, we can substitute t = 1 second into the equation:
s = -16(1)^2 + vo(1) + so
s = -16 + vo + so

Since the initial distance above the ground, so, is 0, the equation simplifies to:
s = -16 + vo

So, the ball is initially 32 feet above the ground, and after 1 second, it will be:
s = -16 + 32
s = 16 feet above the ground.

3. How long will it take to hit the ground?
To find the time it takes for the ball to hit the ground, we need to set s = 0 in the equation and solve for t:
-16t^2 + vot + so = 0

Since the initial distance above the ground, so, is 0, the equation simplifies to:
-16t^2 + vot = 0

We can factor out a common term of t from the equation:
t(-16t + vo) = 0

This equation will be true if either t = 0 or -16t + vo = 0.

If t = 0, it means the ball hasn't been thrown yet, so it doesn't count as hitting the ground. Therefore, we disregard this solution.

If -16t + vo = 0, we can solve for t:
-16t + vo = 0
16t = vo
t = vo/16

So, it will take t = vo/16 seconds for the ball to hit the ground.

4. What is the maximum height of the ball? Show the work.
To find the maximum height of the ball, we need to determine the vertex of the parabolic function s = -16t^2 + vot + so. The vertex represents the highest point on the parabola.

The formula for the x-coordinate of the vertex of a parabola in the form y = ax^2 + bx + c is given by:
x = -b/2a

In our case, the equation relates s (y-coordinate) to t (x-coordinate), so:
t = -vo/(2*(-16))
t = -vo/(-32)
t = vo/32

To find the maximum height, we substitute this value of t into the equation:
s = -16(vo/32)^2 + vo(vo/32) + so
s = -16(vo^2/1024) + vo^2/32 + so
s = (vo^2/32) - (vo^2/64) + so
s = vo^2/64 + so

Since the initial distance above the ground, so, is 0, the equation simplifies to:
s = vo^2/64

So, the maximum height of the ball is vo^2/64 feet.

To find the function that describes the problem, we start with the given information. We are told that the initial velocity (vo) of the ball is 32 feet per second and the initial distance above the ground (so) is 0. We also know that the equation for the distance above the ground (s) as a function of time (t) is given by:

s = 16t^2 + vot + so

Plugging in the values we have, the function becomes:

s = 16t^2 + 32t + 0

Simplifying, the function is:

s = 16t^2 + 32t

To find the height of the ball above the ground after 1 second, we substitute t = 1 into the function:

s = 16(1)^2 + 32(1) = 16 + 32 = 48 feet

So, the ball will be 48 feet above the ground after 1 second.

To find out how long it will take for the ball to hit the ground, we set s = 0 in the function:

0 = 16t^2 + 32t

We have a quadratic equation, so we can solve it by factoring or using the quadratic formula. In this case, we can factor out a common factor of 16t:

0 = 16t(t + 2)

This equation is satisfied when either 16t = 0 or t + 2 = 0. Solving these equations gives us:

t = 0 or t = -2

Since time cannot be negative in this context, we discard t = -2. Therefore, the ball will take 0 seconds to hit the ground, which means it hits the ground instantly.

To find the maximum height of the ball, we can use the vertex formula. The vertex formula states that for a quadratic function in the form f(x) = ax^2 + bx + c, the x-coordinate of the vertex is given by -b/2a, and the y-coordinate of the vertex is the value of the function at that x-coordinate.

In our case, the quadratic function is s = 16t^2 + 32t. Comparing it to the vertex form, we have a = 16, b = 32, and c = 0.

The x-coordinate of the vertex is given by -b/2a:

x = -32 / (2 * 16) = -32 / 32 = -1

The y-coordinate of the vertex is the value of the function at that x-coordinate:

s = 16*(-1)^2 + 32*(-1) = 16 - 32 = -16

Therefore, the maximum height of the ball is -16 feet.

Please note that the negative sign indicates that the maximum height is below the starting point, which in this case is the ground. So, the ball does not reach a maximum height above the ground.