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- Thread starter NINHARDCOREFAN
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I meant 0 = 54-9.8T

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enigma

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Always keep your units in your calculations. It makes it easier to troubleshoot.

Originally posted by NINHARDCOREFAN

For problem 11(please see the pdf file), I did this:

0 =54- 10t

t=5.51

Initial velocity is not 54 m/s, but it washes out in this part, because you know both velocities and can subtract them.

x =54*5.51-XXX*9.8*5.51^{2}

x = 148.8

What did I do wrong?

The initial velocity does not wash in this equation, and you missed a 1/2 term in the acceleration portion.

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enigma

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Originally posted by NINHARDCOREFAN

V= Vi + GT

X= Xi + Vi*T+.5*G*T^{2}

0 = 54 - 10t

t=5.51

x = 54*5.51-9.8*5.51^{2}

Look again, carefully NIN.

54m/s = 108m/s - 9.8m/s

The two like terms can be subtracted, giving you what you got for the first line.

0 = 54m/s - 9.8m/s

That doesn't mean that the initial velocity is now 54m/s.

For the second line, the 1/2 term is

x = 0m + (

- #6

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so I just use the original velocity? I meant to put 1/2 in the equation, it was a typo.

- #7

enigma

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If you think to calculus, the x= equation is merely the integral of the first equation. Since the Vi term is a constant, it just gets a t added to it, but it doesn't change itself.

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Thanks a lot.

- #9

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21.6(squared)= 16.2m/s(squared)-2*9.8m/s(x-10.4m)

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HallsofIvy

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Solve your velocity equation (that you got earlier) equal to 0 to find t when that happens, then plug that t into your height function.

Since velocity is the derivative of height, this is the same as the more general calculus rule: to find max or min of a function, set it's derivative equal to 0.

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