Question

✰✰✰✰★Help me derive this equation★✰✰✰✰
At point O on its path, a projectile has speed V and is travelling at an angle α above the horizontal. Derive the equation of the trajectory of the projectile in the form below

Answer 1

\[y=xtan(\alpha)-\frac{gx^2}{2V^2}(1+\tan^2(\alpha))\]
where x- and y- axes pass through O and are directed horizontally and vertically upwards respectively.

Answer 2

Although I'm good at projectile problems, but this messes me up. How do you start?

Answer 3

The left side:\[x=v_{x}t\]
\[v _{x}=vcos \alpha\]
\[xtan \alpha=v (\cos \alpha )t(\sin \alpha)/\cos \alpha\]
\[=vtsin \alpha=v _{y}\]

Answer 4

*v(y)t

Answer 5

I don't even understand what you're doing

Answer 6

x=v(x)t as distance= speed*time
The third line is replacing in x with v(x)t=vcos(a)t
tanx=sinx/cosx
The 3rd line simplifies to the 4th line, and [vsin(a)]t=[v(y)]t=y

The right side thing is with gravity added in

Answer 7

The right side:
acceleration(y)= -g
v(y)=-gt
y=-0.5g t^2
This is from calculus. Do you understand this?

Answer 8

Anyway, adapting the last formula to suit your question;
\[t^2=\frac{x^2+x^2 \tan^2( \alpha)}{v^2}\]
\[t^2 v^2=x^2(1+ \tan^2( \alpha))\]
\[tv=\sqrt{x^2(1+\tan^2 (\alpha))}= \sqrt{x^2}\sqrt{1+ \tan^2(\alpha)}\]
\[tv=x \sqrt{1+\tan^2 \alpha}\]

Answer 9

tana=v(y)/v(x) (as tanx= opposite/adjacent side)

Answer 10

\[\frac{tv}{x}= \sqrt{1+ \frac{v _{y}^2}{v _{x}^2}}\]
multiplying the right side by v(x)/(v(x)...

Answer 11

\[u_y=usinx \\ \\ u_x=ucosx\]
\[u_x=v_x \\ \\ u_x=\frac{x}{t}\]
\[t=\frac{x}{u_x}\]
From equation
\[y=u_yt-\frac{1}{2}g t^2\]
\[y=(usinx)t-\frac{1}{2}g t^2\]
\[y=usin(x)(\frac{x}{u_x})-\frac{1}{2}g(\frac{x^2}{u_x^2}) \\ \\ y=\cancel{u}\sin(x)(\frac{x}{\cancel{u}\cos(x)})-\frac{1}{2}g\frac{x^2}{u^2\cos^2x} \\ \\ y=xtan(x)-\frac{1}{2}g\frac{x^2}{(u^2\cos^2x)}\]
I got until here

Answer 12

\[tv/x=v/v _{x}\]
\[t=x/v _{x}\]
It seems that I've been working backwards- but trace back my steps and you'll get to that equation.

Answer 13

I think i got it
\[y=xtan(\alpha)-\frac{gx^2}{2u^2\cos^2\alpha}\]
\[y=xtan(\alpha)-\frac{gx^2}{2u^2}(sec^2(\alpha))\]
\[y=xtan(\alpha)-\frac{gx^2}{2u^2}(1+tan^2(\alpha))\]
YES

Answer 14

I'm glad you got it in spite of my cryptic answering.

Answer 15

Thanks to walter lewin too :D
http://www.youtube.com/watch?v=0OV1E9M9Kkc

Answer 16

Odd how omni-pervasive he is: I was considering offering that link by means of an answer..