Question

an object is launched at time t = 0 from a height h above the ground with an initial velocity v0 at an angle θ0 below the horizontal, as shown in the figure. with respect to the indicated coordinate system, which of the following is a correct expression of the position vector (vec{r}) as a function of t for the object after it is launched?
(a) (vec{r}=(v_{0}t)hat{i}+(h - \frac{1}{2}gt^{2})hat{j})
(b) (vec{r}=(v_{0}tcos\theta_{0})hat{i}+(h - v_{0}tsin\theta_{0}-\frac{1}{2}gt^{2})hat{j})
(c) (vec{r}=(v_{0}tcos\theta_{0})hat{i}+(h + v_{0}tsin\theta_{0}-\frac{1}{2}gt^{2})hat{j})
(d) (vec{r}=(-\frac{1}{2}gt^{2}+v_{0}tcos\theta_{0})hat{i}+(h - v_{0}tsin\theta_{0}-\frac{1}{2}gt^{2})hat{j})

Explanation:

Step1: Analyze horizontal motion

In the horizontal ($x$ - direction), there is no acceleration ($a_x = 0$). The initial horizontal - velocity component is $v_{0x}=v_0\cos\theta_0$. Using the kinematic equation $x = x_0+v_{0x}t+\frac{1}{2}a_xt^2$, with $x_0 = 0$ and $a_x = 0$, we get $x = v_0t\cos\theta_0$.

Step2: Analyze vertical motion

In the vertical ($y$ - direction), the initial vertical - velocity component is $v_{0y}=v_0\sin\theta_0$ (downward, so positive in our coordinate system where down is positive $y$), the initial height is $y_0 = H$, and the acceleration is $a_y = g$. Using the kinematic equation $y=y_0 + v_{0y}t+\frac{1}{2}a_yt^2$, we have $y=H + v_0t\sin\theta_0-\frac{1}{2}gt^2$.

Step3: Write position - vector

The position vector $\vec{r}=x\hat{i}+y\hat{j}$. Substituting the expressions for $x$ and $y$ from above, we get $\vec{r}=(v_0t\cos\theta_0)\hat{i}+(H + v_0t\sin\theta_0-\frac{1}{2}gt^2)\hat{j}$.

Answer:

C. $\vec{r}=(v_0t\cos\theta_0)\hat{i}+(H + v_0t\sin\theta_0-\frac{1}{2}gt^2)\hat{j}$