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- anonymous

A ball is thrown vertically upward from the top of a 100-foot tower, with an initial velocity of 20 ft/sec. Its position function is s(t) = –16t2 + 20t + 100. What is its velocity in ft/sec when t = 1 second?
–12
–44
100
–32

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- anonymous

- chestercat

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- carlyleukhardt

C. 100

- anonymous

how did you get that?

- anonymous

If i plug t=1 into the equation I get 104, which isn't an answer

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- anonymous

- anonymous

- carlyleukhardt

huh?

- anonymous

how did you get that answer? can you please explain?

- anonymous

It was wrong @carlyleukhardt

- Photon336

@graze65 you need to visualize this first with a figure

- Photon336

|dw:1443474354027:dw|

- Photon336

The height of our tower is 100 ft and the ball was thrown at an initial velocity of 20ft/second
another thing to add that equation gives you the position not the velocity.
32.1 ft/second is the acceleration due to gravity and that's downward.

- anonymous

so, what do I do next? Do I use the equation for anything? I know I can use the final velocity= initial + (acceleration x speed)

- Photon336

I came up with two possible solutions I'll show you both of them hopefully this will give us the answer

- Photon336

This gives is the equation for the position with respect to time. but we're asked for the velocity with respect to time.
\[s(t) = -16t ^{2} +20t + 100 \]
if we take the derivative, differentiate this position with respect to time we get velocity
\[\frac{ D(s) }{ dT } = -32t + 20 \]
we plug in one second
Vf = we get -32(1) + 20 = -12 ft/second
Second way.
we use the formula
\[V_{o} + at = V_{f}\]
we convert 9.8 meters/second squared to feet per second squared and get 32.1 ft/second squared for the acceleration. the acceleration is pointing down.
20ft/second - 32.1ft/second squared (1second) = -12 ft/second squared = vf
the negative means the velocity is going down so our ball/object is going downwards.

- Photon336

you notice something though, when i took the derivative i got the same kinematic formula.

- anonymous

OHHHHHH I get it now

- anonymous

thank you so much! gonna medal and fan you

- Photon336

no problem :) @graze65 have you taken calculus I? this is why derivatives come in handy.

- anonymous

I'm in calculus now :) this was a calc problem.

- Photon336

\[\frac{ dS }{ dT } = \frac{ \Delta position }{ \Delta time } = velocity\]

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