Projectile Motion using Calculus

Question

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.

Answer 1

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 $$t\approx 4.68 sec$$ $$ and,~~u\approx 85.43$$

Answer 2

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