Question

a rock is thrown from a height $h_0$ with an initial speed $v_0$ at an angle $\theta$ above the horizontal. which of the following expressions represents the time the rock takes to reach the ground after the rock is thrown?
a $\frac{2v_0sin\theta}{g}$
b $\frac{v_0sin\theta}{g}$
c $\frac{v_0sin\theta - sqrt{v_0^{2}sin^{2}\theta + 2gh_0}}{g}$
d $\frac{v_0sin\theta + sqrt{v_0^{2}sin^{2}\theta + 2gh_0}}{g}$

Explanation:

Step1: Analyze vertical - motion equation

The vertical - displacement equation for an object in free - fall is $y - y_0=v_{0y}t-\frac{1}{2}gt^{2}$, where $y - y_0=-h_0$ (negative because the final position is below the initial position), $v_{0y} = v_0\sin\theta$. So, $-h_0=v_0\sin\theta t-\frac{1}{2}gt^{2}$.

Step2: Rearrange to quadratic form

Rearrange the equation to $\frac{1}{2}gt^{2}-v_0\sin\theta t - h_0 = 0$. For a quadratic equation $ax^{2}+bx + c = 0$ (here $x = t$, $a=\frac{1}{2}g$, $b=-v_0\sin\theta$, $c = - h_0$), the quadratic formula is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

Step3: Substitute values into quadratic formula

Substitute $a=\frac{1}{2}g$, $b=-v_0\sin\theta$, $c = - h_0$ into the quadratic formula:
\[

$$\begin{align*} t&=\frac{-(-v_0\sin\theta)\pm\sqrt{(-v_0\sin\theta)^{2}-4\times\frac{1}{2}g\times(-h_0)}}{2\times\frac{1}{2}g}\\ &=\frac{v_0\sin\theta\pm\sqrt{v_0^{2}\sin^{2}\theta + 2gh_0}}{g} \end{align*}$$

\]
We take the positive root because time $t>0$. So $t=\frac{v_0\sin\theta+\sqrt{v_0^{2}\sin^{2}\theta + 2gh_0}}{g}$.

Answer:

D. $\frac{v_0\sin\theta+\sqrt{v_0^{2}\sin^{2}\theta + 2gh_0}}{g}$