Projectile Motion using cos and sin theta??? Golfball is struck to clear a tree 20m away and 6m high at an angle of elevation of 40degrees. Find the speed of the ball when it leaves the ground.
I've created my displacement equation with i and j but i am so lost no clue what to do.
$$v(t) = vt \cos \theta(i) + vt \sin \theta - gt^2/2) (j)$$
can anyone help a brother out?
 A: So you're solving for $v_o$, right?
Okay. Break the initial vector up into its component vectors $v_{oy}$ and $v_{ox}$. The $x$-component of a vector $V$ at an angle $\theta$ degrees above the horizontal is always $V \cos\theta$, and the $y$-component is always $V \sin\theta$. So your $x$- and $y$- components will be $v_{ox}=v_o\cos\theta$ and $v_{oy}=v_o\sin\theta$.
Okay, so you have your initial velocity. Now look at the problem in the $x$- and $y$- directions. You have an $x$ displacement of $6 \text { meters}$ and a $y$ displacement of $20 \text { meters}$. For $x$, speed will remain constant - if there's no drag. So $v_{ox}=v_{fx}$. Using the equation
$$x=\frac{1}{2}(v_f+v_o)t$$
we can simplify it to
$$x=v_{ox}t$$
but we know that $x=20$, so, re-arranging the equation,
$$t=\frac{20}{v_o\cos\theta}$$
Now for the $y$ direction. We know that $v_f$ will not equal $v_o$ if $a$ ($g$) is constant. So we can use the equation
$$y=v_{oy}t+\frac{1}{2}gt^2$$
substituting in for $t$ and $v_{oy}$, we get
$$y=v_o\sin\theta\left(\frac{20}{v_o\cos\theta}\right)+\frac{1}{2}g\left(\frac{20}{v_o\cos\theta}\right)^2$$
Finally, we know that $y=6$ and $\theta=40$, so we can write
$$6=v_o\sin40\left(\frac{20}{v_o\cos40}\right)+\frac{1}{2}g\left(\frac{20}{v_o\cos40}\right)^2$$
Now solve for $v_o$!
