Question

Assuming there's no substitution, could someone help me out on this? It's part of a Physics problem, but all i'm concerned with at the moment is the algebra. (Problem below shortly.)

Answer 1

Going from \[-h = v _{2}t _{2}-\frac{ 1 }{ 2 }g t ^{2}\]to\[t _{2}=\frac{ v _{2}\pm \sqrt{v _{2} ^{2}+2gh}}{ g }\]
I just can't figure out how to get from point a to point b, It seems like it's impossible to get rid of t to entirely one term standing by itself without either substitution or actually trying to factor that being involved. Anybody got an idea?

Answer 2

are you sure you entered the original equations correctly?

Answer 3

Yes, i'm entirely sure. You do have to end up factoring, though. You make it into a quadratic like so: (One sec)

Answer 4

I just figured out the first part, now caught on something else. Starting from the original:\[-h = vt- \frac{ 1 }{ 2 }g t ^{2}\]\[\left(- \frac{ 1 }{ 2 }g t ^{2} + vt = -h \right) \div - \frac{ g }{ 2 }\]\[t ^{2}-\frac{ 2vt }{ g } = \frac{ -2h }{ g } \] From here, add half of the second term's constants on both sides and square that resulting term:\[t ^{2}-\frac{ 2vt }{ g }+\frac{ v ^{2} }{ g ^{2} } = -\frac{ 2h }{ g }+\frac{ v ^{2} }{ g ^{2} }\]Factor the left side:\[(t - \frac{ v }{ g }) = \sqrt{-\frac{ 2h }{ g }+\frac{ v ^{2} }{ g ^{2} }}\]Isolate t. But here I get caught at a new portion.

Answer 5

EDIT: How do I get from\[t = \frac{ v }{ g } \pm \sqrt{\frac{ 2h }{ g }+\frac{ v ^{2} }{ g ^{2} }}\]to\[t = \frac{ v \pm \sqrt{2gh + v ^{2}}}{ g }\]?

Answer 6

g=√g^2

Answer 7

Did you multiply by g/g on both sides and do that, or something else?

Answer 8

something like that , yes

Answer 9

Nevermind, just figured it out. My bad. Thanks.

Answer 10

\[ax^2+bx+c=0\]\[x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]\[(x-x_1)(x-x_2)=0\]


\[-h = v _2t-\tfrac12g t ^{2}\\\tfrac12gt^2-v_2t+h=0\]
\[t=\frac{-(-v_2)\pm\sqrt{(-v_2)^2-4\times\tfrac 12g\times h}}{g}\\\quad=\frac{v_2\pm\sqrt{v_2^2-2gh}}{g}\]

Answer 11

its just quadratic formula