s = -16t^2 + v t + s: when will it hit the ground, etc?

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I think I almost understand but I have run into another problem:

Suppose you throw a baseball straight up at a velocity 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 = -16t^2 + v t + s

A) The ball will be how high above the ground after 1 second

B) How long will it take to hit the ground?

C) How high does the ball go?

How do I do a negative for A= -16? I am not sure I understand? The next question on my sheet is I have to graph this. Since it is a parabola the curve would look like a frown correct? I know this is more than one question but I am really struggling with this class
 
What do you mean by "doing a negative for A = -16"? Please clarify. (It would probably help if you listed out the steps you have tried, and stated specifically at what point you are stuck.)

Thank you.

Eliz.
 
the same

ok it looks like prepage and I are in the same class!! I have only gotten about this far with this problem 0= -16t^2 + 32t + 0 if t=1 than

after 1 second the ball will be 16 ft in the air Correct?
How do I write the equation to tell how high the ball will go and how do I tell when the ball will come back down?
 
can somebody just check my work?

can somebody just check my work? I think I got it but I don't know. I am unsure if the answers are supposed to be negative or positive

a)
st = -16 + 32 + 0
s(1) = -16 (1)^2 + 32(1)
1s= -16 + 32
1s =16

so the ball would be 16 feet in the air after 1 second

c)
t= -b/2a = -32/2(-16) = -1 or 1

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

vertex would be 1,-16
 
Suppose you throw a baseball straight up at a velocity 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 = -16t^2 + v t + s

A) The ball will be how high above the ground after 1 second

B) How long will it take to hit the ground?

C) How high does the ball go?
Click to expand...

Hi, Brandie,

Here’s what’s going on: Consider the various terms of the function s(t) = -16t^2 + v t + s; v = 32, s = 0

The “s” at the end on the right side of the function represents the initial position of the object; in this case, s = 0 means the object is starting at ground level, which we are calling “0”.

The “v” term is the initial velocity of the object, in this case v = 32 ft/sec upwards, and we are defining “up” to be the positive direction, so “32” is positive.

The coefficient of the “t^2” term is “-16.” This number comes from one-half of the acceleration due to gravity, or (1/2)*(32 ft/sec^2) = 16 ft/sec^2. However, we are defining the downward direction to be negative (since “up” is positive), so we use –16 ft/sec^2.

You’ve accurately identified this as a parabola that opens downward, so its vertex will be its highest point. The x-axis is time (t) in seconds, and the y-axis is height (s) in feet. Therefore the vertex represents the highest point the object will reach.

c)
t= -b/2a = -32/2(-16) = -1 or 1

s(1) = -16 (1)^2 +32(1)
1s= -16 + 32
Click to expand...

s(1) = 16 (not –16)

The vertex is at t = 1 second. The object is at 16 feet then.

Setting s(t) = 0 is how we’d find when the object is back at “0” (ground level).

0 = -16t^2 + 32t = -16t(t – 2)

What are the solutions for t?
 
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