Question

3). a rock is thrown with an initial vertical velocity component of 30 m/s and an initial horizontal velocity component 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?
b. what is the total time the rock is in the air?
c. how far will the rock land from where it was thrown?

Explanation:

Step1: Analyze horizontal - vertical motion

In projectile motion, horizontal velocity ($v_x$) is constant as there is no horizontal acceleration ($a_x = 0$). At the top - most point of the vertical motion, the vertical velocity ($v_y$) is 0. Given $v_{0x}=40$ m/s and $v_{0y}=30$ m/s, $v_{x,\text{at top}} = 40$ m/s and $v_{y,\text{at top}}=0$ m/s.

Step2: Calculate time of flight for vertical motion

Use the kinematic equation $v = v_0+at$ for vertical motion. When the rock lands at the same height, the displacement in the y - direction $y - y_0 = 0$. The kinematic equation $y - y_0=v_{0y}t-\frac{1}{2}gt^2$. Since $y - y_0 = 0$, we can also use $v = v_0+at$. At the maximum height, $v_y = 0$ and $v_y=v_{0y}-gt_{up}$, so $t_{up}=\frac{v_{0y}}{g}$. The total time of flight $t=\frac{2v_{0y}}{g}$. Given $v_{0y}=30$ m/s and $g = 9.8$ m/s², $t=\frac{2\times30}{9.8}\approx6.12$ s.

Step3: Calculate horizontal range

The horizontal range $R = v_{0x}t$. We know $v_{0x}=40$ m/s and $t\approx6.12$ s. So $R = 40\times6.12 = 244.8$ m.

Answer:

a. $v_{x,\text{at top}} = 40$ m/s, $v_{y,\text{at top}} = 0$ m/s
b. $t\approx6.12$ s
c. $R = 244.8$ m