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anonymous
 one year ago
A ball is thrown vertically upward from the top of a 100foot 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
anonymous
 one year ago
A ball is thrown vertically upward from the top of a 100foot 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
 one year ago
Best ResponseYou've already chosen the best response.0how did you get that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If i plug t=1 into the equation I get 104, which isn't an answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did you get that answer? can you please explain?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It was wrong @carlyleukhardt

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@graze65 you need to visualize this first with a figure

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443474354027:dw

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1The 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
 one year ago
Best ResponseYou've already chosen the best response.0so, 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
 one year ago
Best ResponseYou've already chosen the best response.1I came up with two possible solutions I'll show you both of them hopefully this will give us the answer

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1This 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
 one year ago
Best ResponseYou've already chosen the best response.1you notice something though, when i took the derivative i got the same kinematic formula.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OHHHHHH I get it now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much! gonna medal and fan you

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1no problem :) @graze65 have you taken calculus I? this is why derivatives come in handy.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm in calculus now :) this was a calc problem.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ dS }{ dT } = \frac{ \Delta position }{ \Delta time } = velocity\]
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