Projectile Motion using Calculus 
A ball is thrown from a height of 6 ft at an angle of 60 degrees with the horizontal. If the ball lands $200$ ft away, what was the initial speed? 

I know of the equation $$x_{\text{range}}= \frac{v_o^2\sin a}{g}$$ but since the ball starts at an initial height, I don't know if it can applied in this scenario. If not, does anyone know how to do this using vector valued functions? Much appreciated. 
 A: It is solvable using only those projectile motion's formula,
Let, the downward direction be $(+)ve$ and upward be $(-)ve$.
As,the ball has crossed 6 ft distance vertically in the downward direction.so it is positive.but as it was thrown $60^{\circ}$ upward.so the vertical component of its initial velocity is negative.And as $g$ is working in downward.so it is positive too.
so we can write,
$$y=-u\sin 60^{\circ} .t+\dfrac{1}{2}g.t^2$$
$$\implies 6=-u\sin 60^{\circ} .t+\dfrac{1}{2}*32.15*t^2..........(1) $$
Again we get,for horizontal component of the velocity,
$$x=u\cos 60^{\circ}.t............(2)$$
If you solve these two equations,then you will get
$$t\approx 4.68 sec$$
$$ and,~~u\approx 85.43$$ 
A: Fix the ground as the datum
$y = y_0 +v_0sin\theta t - \frac{1}{2}gt^2$
$ R  = vcos\theta. t$
$t = \frac{200}{v_0cos\theta}$
Substituting the value of t in 1, we get
$0 = 6 + \frac{v_0sin\theta. 200}{v_0cos\theta} -\frac{1}{2}g \frac{200^2}{v_0^2cos^2\theta}$
Solving for $v_0$
$200tan\theta + 6 =\frac{1}{2}g \frac{200^2}{v_0^2cos^2\theta}$
$200\sqrt{3}+ 6 =\frac{1}{2}g \frac{40000.4}{v_0^2}$
$v_0^2 = \sqrt{\frac{160000.(32.15)}{2.352.4}}$
$v_0 = 85.43$ft/s
