Question

3). a rock is thrown with an initial vertical velocity component of 30 m/s and an initial horizontal velocity component of 40 m/s. assume the rock lands at the same height it starts with. a. what will the velocity components of both the x & y direction be when the rock reaches the top of its path? vx, at top=__ vy, at top=__ b. what is the total time the rock is in the air?

Explanation:

Step1: Analyze horizontal - motion

In horizontal direction, there is no acceleration ($a_x = 0$) for a projectile motion. So the horizontal - velocity component remains constant throughout the motion. Given $v_{0x}=40\ m/s$, at the top of the path, $v_{x,\text{at top}}=v_{0x}$.
$v_{x,\text{at top}} = 40\ m/s$

Step2: Analyze vertical - motion

At the top of the path of a projectile, the vertical - velocity component is zero. Because the rock stops moving upward and is about to start moving downward. So $v_{y,\text{at top}} = 0\ m/s$.

Step3: Calculate total time of flight

We use the kinematic equation for vertical motion $v = v_0+at$. When the rock lands at the same height it starts with, the displacement in the y - direction $y - y_0 = 0$. The kinematic equation $y - y_0=v_{0y}t-\frac{1}{2}gt^2$ can be used. Also, we can use the equation $v = v_0+at$. At the end of the motion (when it lands at the same height), $v_y=-v_{0y}$. Using $v_y = v_{0y}-gt$, and setting $v_y=-v_{0y}$, we get $-v_{0y}=v_{0y}-gt$.
Solving for $t$:
\[

$$\begin{align*} gt&=2v_{0y}\\ t&=\frac{2v_{0y}}{g} \end{align*}$$

\]
Given $v_{0y} = 30\ m/s$ and $g = 9.8\ m/s^2$, then $t=\frac{2\times30}{9.8}\approx6.12\ s$

Answer:

a. $v_{x,\text{at top}} = 40\ m/s$, $v_{y,\text{at top}} = 0\ m/s$
b. $t\approx6.12\ s$