Getting the sines and cosines right
Are you unsure about which of the components of the launch velocity force is v0sinθ and which is v0cosθ? If so
- You are not alone
- You can fix it once and for all
To ‘fix’ it you do need to know a few basic things about the sine and the cosine; specifically:
![\[ \sin {0} = \cos {(\pi/2)} =0 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \cos {0} = \sin {(\pi/2)} =1 \]](mastermathpng-0.png)
Now just appeal to the ‘special cases’ trick (Guideline 0.10). You can see that the horizontal (x) component of the velocity has to be vx=v0cosθ because
- When θ=0 (the launch is ‘horizontal’) this expression implies vx=v0, as it has to
- When θ=π/2 (the launch is straight up) this expression implies vx=0, as it has to.
Please learn this now...it will save you (and us) much pain.