TBR Physics I - Page 25

[bluishgreen]

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I am mind fkd after seeing this equation(after paragraph 2) change so many times. And the second paragraph doesn't make any sense to me. can someone please explain it?

 

[Catburr]

Can you be more specific about what confuses you? It's ok to quote the books.

The paragraph (at least in my book, with the blue cover, 2009 edition that I bought late last year) is about applying algebra to factor out time in the translational motion equation y = (initial velocity)(time) + (1/2)at^2.

Algebra, and the logic that y = 0 when the ball is on the ground.

 

[bluishgreen]

The question which asks to find T_f_. I don't get it because it says to use equation 1.18 ∆y=v_oy ∆t+1/2 a_y (∆t^2) . And how the heck does this equation change from this to t=2v_o/g ? I feel like this paragraph is all riddled up.

 

[Blaughable]

So the question is asking when will the ball return back to the ground. Well that is when will the ball be at y=0.

Using Yf = Yo + Vot + 1/2at^2
where Yo=0, Yf=0

we get the equation:
0 = Vot + 1/2at^2

You can factor a t out of the right side and get:
0 = t(Vo+1/2at)

this equation can be true when either t=0 (before it is launched) or when what is inside the bracket equals 0:

Vo+1/2at = 0
t = -2Vo/a (a = -g)
t = 2Vo/g

 

[bluishgreen]

this equation can be true when either t=0 (before it is launched) or when what is inside the bracket equals 0:

Vo+1/2at = 0
t = -2Vo/a (a = -g)
t = 2Vo/g

I don't get this. Can you please explain the math used here?

 

[Blaughable]

I don't get this. Can you please explain the math used here?

NO problem:
Vo + 1/2at = 0

1.) Subtract Vo from both sides
(Vo + 1/2at)-Vo = 0-Vo
1/2at = -Vo

2.)Multiply by 2
2*(1/2at) = 2*(-Vo)
at = -2Vo

3.) divide by acceleration
(at)/a = (-2Vo)/a
t = -2Vo/a

 

[bluishgreen]

thanks soooo much.