Acceleration#5 - Derivation

OK, first I want to bring back 4 equations from last time…

sy=(1/2)at²+v_yt+s₀
s_x=v_xt
v_x=vcosθ
v_y=vsinθ

…and a trigonometric identity.

1/(cos2θ)=tan2θ+1

Now we can derive a new equation to fit what we’ll do today.

s_x=v_xt
↓
t=s_x/v_x
↓
t=s_x/(vcosθ)

sy=(1/2)at²+v_yt+s₀

Since s₀ is the origin, it equals 0 and would not need to be included.

s_y=(1/2)at²+v_yt
↓
0=(1/2)at²+v_yt-s_y
↓
0=(1/2)at²+vsinθt-s_y
↓
0=(1/2)a(s_x/(vcosθ))²+vsinθ(s_x/(vcosθ))-s_y
↓
0=((as_x²)/v²)(1/cos²θ)+s_x(sinθ/cosθ)-s_y
↓
0=(as_x²/v²)(tan²θ+1)+s_xtanθ-s_y
↓
0=as_x²/v²•tan²θ+s_xtanθ+ as_x²/v²-s_y