How Do You Isolate 'v' in a Projectile Range Equation with Initial Height?

[xovangam]

given this formula for the range of a projectile when the initial height (y) is not zero:

d = (v * cos(a) / g) * (v * sin(a) + sqrt((v * sin(a))[SUP]2[/SUP] + 2 * g * y)

in a project I'm working on i need to compute 'v' given 'd', 'a', 'y' and 'g', so i would like to rewrite this equation in terms of 'v' - my trig identities aren't so hot, could someone possibly help me to express this equation in terms of 'v' instead of 'd' ?

TIA

*er - this was supposed to post in General Math...oops. i don't know how to move it there w/o crossposting..

 

[micromass]

Hi xovangam!


So the equation is

$$d=\frac{v\cos(a)}{g}(v\sin(a)+\sqrt{v^2\sin^2(a)+2gy})$$

Right?

The square root is the most important problem here. So you'll first going to need to rewrite your equation as

$$\frac{gd}{v\cos(a)}-v\sin(a)=\sqrt{v^2\sin^2(a)+2gy}$$

and square both sides. You'll end up with a quadratic equation in v.

 

[xovangam]

is this on the right track ?

((gd/v*cos(a)) - v*sin(a))^2 = v^2*sin(a)^2 + 2gy

A = gd/v*cos(a)
B = v*sin(a)

(A - B)(A - B)
= A^2 - AB - AB + B^2
= A^2 - 2AB + B^2

(gd/v*cos(a))^2 - 2*(gd/v*cos(a))*(v*sin(a)) + (v*sin(a))^2 = v^2*sin(a)^2 + 2*g*y

=> (gd)^2/(v*cos(a))^2 - (2*gd*v*sin(a))/(v*cos(a)) + v^2*sin(a))^2 = v^2*sin(a)^2 + 2*g*y

=> (gd)^2/(v^2*cos(a)^2) - (2*gd*sin(a))/(cos(a)) = 2*g*y

how do i get v^2*cos(a)^2 out of the bottom of the fraction on the left, multiply both sides by v^2*cos(a)^2 ? it seems like I'm going to end up with 'B' = 0 in the simplified equation ? (i.e. Av^2 + Bv + C)

 

[micromass]

Looks good. So now have something of the form

$$\frac{A}{Bv^2}+C=D$$

This is very easy to solve for v! Indeed:

$$\frac{A}{Bv^2}=D-C$$

Thus

$$\frac{Bv^2}{A}=\frac{1}{D-C}$$

Hence

$$v^2=\frac{A}{B(D-C)}$$

 

[xovangam]

cool. so

A = (g*d)^2
B = cos(a)^2
C = -(2*g*d*sin(a))/(cos(a))
D = 2*g*y

and thus

v^2 = (g*d)^2/(cos(a)^2*(2*g*y + (2*g*d*sin(a))/cos(a))

?

 

[micromass]

Seems right!

 

[xovangam]

and can i further simply that to:

v = (g*d)/cos(a) * sqrt(2 * g * d * tan(a) + 2 * g * y)

?

 

[micromass]

If you know that everything is positive, yes.

 

[Ray Vickson]

Hi xovangam!


So the equation is

$$d=\frac{v\cos(a)}{g}(v\sin(a)+\sqrt{v^2\sin^2(a)+2gy})$$

Right?

The square root is the most important problem here. So you'll first going to need to rewrite your equation as

$$\frac{gd}{v\cos(a)}-v\sin(a)=\sqrt{v^2\sin^2(a)+2gy}$$

and square both sides. You'll end up with a quadratic equation in v.

No. You will end up with a quadratic equation in v^2, so if you set y = v^2 you get a quadratic in y.

RGV

 

[micromass]

No. You will end up with a quadratic equation in v^2, so if you set y = v^2 you get a quadratic in y.

RGV

If you check the solution in the previous post, you see that you do end up with a quadratic equation in v. Is the solution wrong somewhere?

 

[Ray Vickson]

The equation you wrote before had the form
A/v - Bv = sort(v^2 + c), so squaring it gives
A^2 /v^2 - 2AB + B^2 v^2 = c + v^2. Multiply through by v^2 to get a quadratic in v^2.

RGV

 

[micromass]

The equation you wrote before had the form
A/v - Bv = sort(v^2 + c), so squaring it gives
A^2 /v^2 - 2AB + B^2 v^2 = c + v^2. Multiply through by v^2 to get a quadratic in v^2.

RGV

And in this case, it was a quadratic. Like I claimed. I don't see your point.

 

[Ray Vickson]

I was responding to the statement that it is quadratic in *v* (not v^2). Of course, that might have been a typo.

RGV

 

[micromass]

I was responding to the statement that it is quadratic in *v* (not v^2). Of course, that might have been a typo.

RGV

Really, excuse me, I don't understand what the confusion is here. If you square both sides of the equation and rework it a bit, then the resulting equation will be quadratic in v, no??
This is certainly not true with every such equation (where you indeed have to make a substitution), but it is true with this equation, and that's all I said...

 

[Ray Vickson]

The equation becomes A^2 - 2AB v^2 + B^2 v^4 = c v^2 + v^4, where B is not equal to +1 or -1, so the v^4 terms do not cancel.

RGV

 

[micromass]

The equation becomes A^2 - 2AB v^2 + B^2 v^4 = c v^2 + v^4, where B is not equal to +1 or -1, so the v^4 terms do not cancel.

RGV

In this case, they do! So in this case, we do have a quadratic equation. (or did I make a mistake anywhere??). I really don't see what the big deal is here.

 

[Ray Vickson]

You are right! Sorry: I was looking at the wrong equation.

RGV

 

[JJacquelin]

In brief :