Question

a projectile is launched horizontally from a cliff top at 18 m/s. determine the velocity components at 1 - second intervals of time. enter a - for left and down. using ( g = 9.8 m/s/s ). enter - for left and down.

Explanation:

Step1: Analyze the horizontal velocity component

In projectile motion, the horizontal velocity \(v_x\) remains constant (assuming no air - resistance). The initial horizontal velocity \(v_{x0}=18\ m/s\). So for all times \(t\), \(v_x = 18\ m/s\).

• At \(t = 1.0\ s\), \(v_{x1}=18\ m/s\) (A)
• At \(t = 2.0\ s\), \(v_{x2}=18\ m/s\) (C)
• At \(t = 3.0\ s\), \(v_{x3}=18\ m/s\) (E)
• At \(t = 4.0\ s\), \(v_{x4}=18\ m/s\) (G)

Step2: Analyze the vertical velocity component

The vertical velocity component is given by the formula \(v_y=v_{y0}+at\). Since the projectile is launched horizontally, \(v_{y0} = 0\ m/s\) and \(a=-g=- 9.8\ m/s^2\) (taking down - ward as negative).

• For \(t = 1.0\ s\):

\(v_{y1}=0+( - 9.8)\times1=-9.8\ m/s\) (B)

• For \(t = 2.0\ s\):

\(v_{y2}=0+( - 9.8)\times2=-19.6\ m/s\) (D)

• For \(t = 3.0\ s\):

\(v_{y3}=0+( - 9.8)\times3=-29.4\ m/s\) (F)

• For \(t = 4.0\ s\):

\(v_{y4}=0+( - 9.8)\times4=-39.2\ m/s\) (H)

Answer:

time (s)	\(v_x(m/s)\)	\(v_y(m/s)\)
\(1.0\)	\(18\)	\(-9.8\)
\(2.0\)	\(18\)	\(-19.6\)
\(3.0\)	\(18\)	\(-29.4\)
\(4.0\)	\(18\)	\(-39.2\)