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Mathematics 19 Online
OpenStudy (arianna1453):

FAN + MEDAL A paper clip is dropped from the top of a 144‐ft tower, with an initial velocity of 16 ft/sec. Its position function is s(t) = −16t^2 + 144. What is its velocity in ft/sec when it hits the ground?

OpenStudy (mathmale):

Let's break this problem into parts (always a good strategy). The clip falls with an initial velocity of 16 ft/sec and drops 144 feet. How long does it take it to reach the ground? sorry for the delay in responding to your last question. y=mx+b is an appropriate form for your answer to that question.

OpenStudy (mathmale):

Hint: What is the formula for velocity, given that its position function is s(t) = −16t^2 + 144?

OpenStudy (arianna1453):

I literally could not tell you, Velocity problems confuse me.

OpenStudy (mathmale):

Important: the velocity function is just the derivative of the position function. if the latter is s(t) = −16t^2 + 144, what's s '(t)?

OpenStudy (arianna1453):

-32t

OpenStudy (mathmale):

that's right. Good. Next, determine how long it will take the clip to reach the ground. How would you do that? Note that you're dealing in distance, measured in feet.

OpenStudy (arianna1453):

mmmm.... I am really not sure. Something to do with time?

OpenStudy (mathmale):

You're to DETERMINE the length of time req'd for the clip to hit the ground if the initial velocity is 16 ft/sec and the initial height is 144 ft.

OpenStudy (arianna1453):

set it equal to 0?

OpenStudy (mathmale):

You are given a formula for the position (height) of the clip, plus you are given a specific initial height. Yes, setting the formula for height = 0 correctly models the clip's hitting the ground.

OpenStudy (arianna1453):

So setting the derivative equal to zero? or the regular equation? since that is what has the initial height

OpenStudy (mathmale):

We have to keep position and velocity separate.

OpenStudy (mathmale):

Please go back to the original problem statement, copy the formula for position, and then paste it here. Then set that equation equal to zero (which indicates that the clip is on the ground).

OpenStudy (arianna1453):

s(t) = −16t^2 + 144=0 t=3 ,-3 But since it cant be negative, the time is just 3. so t=3

OpenStudy (mathmale):

So, you're saying that the clip hits the ground after 3 sec.? That looks reasonable. Repeat the equation for velocity here, please.

OpenStudy (arianna1453):

Plugging 3 into t for the derivative you get : -32t = -32(3) = -96?

OpenStudy (mathmale):

Yes. label that as s '(3) = v(3) = 96. What are the units of measurement, and should your result be positive or neg? why?

OpenStudy (arianna1453):

ft/sec and negative

OpenStudy (arianna1453):

Because it hits the ground?

OpenStudy (mathmale):

Because it's moving downward. Downward is neg here, upward is pos.

OpenStudy (mathmale):

I see you've moved on to your next homework problem (and certainly understand your need to get all of your homework done. Hope our discussion has been helpful.

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