X = 54*5.51-9.8*5.51(squared)

1. Sep 13, 2003

NINHARDCOREFAN

For problem 11(please see the pdf file), I did this:
0 = 54 - 10t
t=5.51

x = 54*5.51-9.8*5.51(squared)
x = 148.8

What did I do wrong?

Note:
V= Vi + GT
X= Xi + Vi*T+.5*G*T(squared)

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Last edited by a moderator: Feb 5, 2013
2. Sep 13, 2003

NINHARDCOREFAN

Mistake

I meant 0 = 54-9.8T

3. Sep 13, 2003

enigma

Staff Emeritus
Re: Problem

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

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

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

4. Sep 13, 2003

NINHARDCOREFAN

yeah, .5*108=54. I didn't forget the 1/2. I don't understand when you say, you have to subract both velocities, because if you do that you get the same velocity

5. Sep 13, 2003

enigma

Staff Emeritus
Re: Problem

Look again, carefully NIN.

54m/s = 108m/s - 9.8m/s2*t

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

0 = 54m/s - 9.8m/s2*t

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

For the second line, the 1/2 term is not on the velocity portion, it's on the acceleration portion!

x = 0m + (108m/s * 5.51s) + (1/2 * 9.8m/s2 * t2)

6. Sep 13, 2003

NINHARDCOREFAN

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

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

7. Sep 13, 2003

enigma

Staff Emeritus
Yep, that's it.

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.

8. Sep 13, 2003

NINHARDCOREFAN

Thanks a lot.

9. Sep 13, 2003

NINHARDCOREFAN

Another question, for problem 13(please see the pdf file) what do i do? I'm getting 0 as an answer
21.6(squared)= 16.2m/s(squared)-2*9.8m/s(x-10.4m)

10. Sep 14, 2003

HallsofIvy

Staff Emeritus
The stone will go up as long as it's velocity is positive, come down when it's velocity is negative. It will be at it's highest point when it's velocity is 0.
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.