Prove this equation for projectile motion

[Winner123]

Homework Statement

An object launched with a speed of vi from a height deltay above the horizontal floor will land a certain horizontal distance deltax from the launch point. the initial speed can be shown to be given by
vi=√(xf-xi)^2g/2(yf-yi)
proof this equation

Relevant Equations

vi=√(xf-xi)^2g/2(yf-yi)

I tried using the formulas x=xi+vit and y=yi+voyt-1/2g(t^2)
I assumed voy would be 0 and I almost arrive to the answer but idk how to get rid of the negative

 

[PeroK]

What did you do and where did you get stuck?

 

[haruspex]

Homework Statement: An object launched with a speed of vi from a height deltay above the horizontal floor will land a certain horizontal distance deltax from the launch point. the initial speed can be shown to be given by
vi=√(xf-xi)^2g/2(yf-yi)

Presumably the initial velocity is horizontal. Did you leave that out or was it omitted in the original?
The given answer is clearly wrong since $y_f-y_i$ is negative.
Please clarify the scope of the square root by using parentheses or, better, using LaTeX.

 

[kuruman]

$\dots$ but idk how to get rid of the negative

Given that the initial velocity is horizontal, you are asked to show that the initial speed can be written as $$v_i=\sqrt{-\frac{g(x_f-x_i)^2}{2(y_f-y_i)}}.$$ If you have already arrived at this expression, there is no negative sign to get rid of because $(y_f-y_i)<0.$

 

[PeroK]

You need a degree in cryptography to be a homework helper these days. Is this something to do with the Generation Z that I've been hearing about?

 

[kuruman]

You need a degree in cryptography to be a homework helper these days. Is this something to do with the Generation Z that I've been hearing about?

idk

 

[haruspex]

Given that the initial velocity is horizontal, you are asked to show that the initial speed can be written as $$v_i=\sqrt{-\frac{g(x_f-x_i)^2}{2(y_f-y_i)}}.$$ If you have already arrived at this expression, there is no negative sign to get rid of because $(y_f-y_i)<0.$

As I posted, the given equation is wrong because it does not have that minus sign, unless your eyes are much better than mine.

 

[kuruman]

As I posted, the given equation is wrong because it does not have that minus sign, unless your eyes are much better than mine.

Indeed it does not. In post #4 I show the equation that OP is supposed to show in readable form. We don't know whether the equation as given to OP has or does not have the minus sign. I suspect that OP got the correct equation but posted what is to be shown without it thinking that no minus sign belongs under a radical ever. Then OP did the algebra correctly, ended up with a minus sign and is asking us how to "get rid" of the minus sign.

Convoluted? Perhaps, but this is what happens when it is drilled in one's head that overall negative signs do not belong under radicals. There is an easy way to find out, so let's wait to see what OP has to say.

 

[haruspex]

Indeed it does not. In post #4 I show the equation that OP is supposed to show in readable form. We don't know whether the equation as given to OP has or does not have the minus sign. I suspect that OP got the correct equation but posted what is to be shown without it thinking that no minus sign belongs under a radical ever. Then OP did the algebra correctly, ended up with a minus sign and is asking us how to "get rid" of the minus sign.

Convoluted? Perhaps, but this is what happens when it is drilled in one's head that overall negative signs do not belong under radicals. There is an easy way to find out, so let's wait to see what OP has to say.

Since the OP's issue seems to be trying and failing to get rid of the minus in his/her own attempt, it seems unlikely that such a minus sign in the given equation was overlooked.

 

[Herman Trivilino]

Although it's rare, some textbook authors write $g=\pm 9.8\ \mathrm{m/s^2}$ with a convention for when to use which sign.

 

[PeroK]

The given answer is clearly wrong since $y_f-y_i$ is negative.

Not if the positive y direction is taken to be downwards.

 

[haruspex]

Not if the positive y direction is taken to be downwards.

from a height delta y above the horizontal floor