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?
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 vertical motion for time - up part

The vertical - motion of the rock is a free - fall motion. At the top of the path, the vertical velocity $v_y = 0$. The initial vertical velocity $v_{0y}=30\ m/s$, and the acceleration $a=-g=- 9.8\ m/s^{2}$. Using the kinematic equation $v = v_0+at$, for the upward motion to the top $0 = 30-9.8t_1$, so $t_1=\frac{30}{9.8}\ s$. The total time of flight $t$ (since the rock lands at the same height) is $t = 2t_1$.
$t=\frac{2\times30}{9.8}\ s\approx6.12\ s$

Step2: Analyze horizontal motion for range

The horizontal motion of the rock is a uniform - motion with constant velocity because there is no acceleration in the horizontal direction ($a_x = 0$). The initial horizontal velocity $v_{0x}=40\ m/s$, and the time of flight $t = 6.12\ s$. Using the equation $x=v_{0x}t$, we can find the horizontal range.
$x = 40\times6.12=244.8\ m$

Answer:

a. $v_{x,\text{at top}} = 40\ m/s$, $v_{y,\text{at top}} = 0\ m/s$
b. $t\approx6.12\ s$
c. $x\approx244.8\ m$