SOLUTION: Solve for t. h= (1/2)gt^2+v0t I was thinking you have to multiply both sides by 2 first but I don't know what to do with the "plus v of t" at the end.

Algebra ->  Equations -> SOLUTION: Solve for t. h= (1/2)gt^2+v0t I was thinking you have to multiply both sides by 2 first but I don't know what to do with the "plus v of t" at the end.      Log On


   

Question 483670: Solve for t. h= (1/2)gt^2+v0t
I was thinking you have to multiply both sides by 2 first but I don't know what to do with the "plus v of t" at the end.
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

h=expr%281%2F2%29gt%5E2%2Bv%5B0%5Dt

As you said, multiply through by 2

2h=gt%5E2%2B2v%5B0%5Dt

0+=+gt%5E2%2B2v%5B0%5Dt-2h

gt%5E2%2B2v%5B0%5Dt-2h=0

That is a quadratic equation, so we use the
quadratic formula:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ 

with x = t, a = g, b = 2v0, and c = -2h

 

t+=+%28-2v%5B0%5D+%2B-+sqrt%28+4v%5B0%5D%5E2%2B8gh%29+%29%2F%282%2Ag%29+

t+=+%28-2v%5B0%5D+%2B-+sqrt%28+4%28v%5B0%5D%5E2%2B2gh%29%29+%29%2F%282%2Ag%29+

t+=+%28-2v%5B0%5D+%2B-+2sqrt%28+v%5B0%5D%5E2%2B2gh%29+%29%2F%282%2Ag%29+

t+=+%282%28-v%5B0%5D+%2B-+sqrt%28+v%5B0%5D%5E2%2B2gh%29%29+%29%2F%282%2Ag%29+



t+=+%28-v%5B0%5D+%2B-+sqrt%28+v%5B0%5D%5E2%2B2gh%29%29+%2Fg+

There are two solutions.  

t+=+%28-v%5B0%5D+-+sqrt%28+v%5B0%5D%5E2%2B2gh%29%29+%2Fg+ and t+=+%28-v%5B0%5D+%2B+sqrt%28+v%5B0%5D%5E2%2B2gh%29%29+%2Fg+

The given equation is for the time when an object projected upward from the
ground is at a certain height h.  It is at that height h once going up and
once again coming down.  The first solution is for the time when the ball
first reaches height h going up, and the second is for the time when it
reaches that same height h coming down.   

Edwin