Question

a projectile is launched horizontally from a cliff top at 16 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. table with time (s), vx (m/s), vy (m/s) rows: 0.0 (16, 0.0), 1.0 (a: blank, b: blank), 2.0 (c: blank, d: blank), 3.0 (e: blank, f: blank), 4.0 (g: blank, h: blank) once satisfied with your entries, tap on check answers. image of projectile motion with points b, d, f, h downward and a, c, e, g rightward

Explanation:

Step1: Analyze Horizontal Velocity

In horizontal projectile motion, there is no horizontal acceleration (assuming no air resistance), so the horizontal velocity \( v_x \) remains constant. The initial horizontal velocity is 16 m/s, so \( v_x = 16 \) m/s for all times.

Step2: Analyze Vertical Velocity

The vertical motion is free - fall motion with acceleration \( g = 9.8\ m/s^2 \) (downward). The formula for vertical velocity in free - fall is \( v_y=v_{y0}+gt \), where \( v_{y0} = 0\ m/s \) (initial vertical velocity is 0 as the projectile is launched horizontally).

For \( t = 1.0\ s \):

\( v_y=0 + 9.8\times1.0=- 9.8\ m/s \) (negative because it is downward)

For \( t = 2.0\ s \):

\( v_y=0+9.8\times2.0 = - 19.6\ m/s \)

For \( t = 3.0\ s \):

\( v_y=0 + 9.8\times3.0=-29.4\ m/s \)

For \( t = 4.0\ s \):

\( v_y=0+9.8\times4.0=-39.2\ m/s \)

Filling the table:

• For \( t = 1.0\ s \):
• \( A = 16 \) (horizontal velocity)
• \( B=-9.8 \) (vertical velocity)
• For \( t = 2.0\ s \):
• \( C = 16 \) (horizontal velocity)
• \( D=-19.6 \) (vertical velocity)
• For \( t = 3.0\ s \):
• \( E = 16 \) (horizontal velocity)
• \( F=-29.4 \) (vertical velocity)
• For \( t = 4.0\ s \):
• \( G = 16 \) (horizontal velocity)
• \( H=-39.2 \) (vertical velocity)

Answer:

time (s)	\( v_x \) (m/s)	\( v_y \) (m/s)
1.0	16	- 9.8
2.0	16	- 19.6
3.0	16	- 29.4
4.0	16	- 39.2

So, \( A = 16 \), \( B=-9.8 \), \( C = 16 \), \( D=-19.6 \), \( E = 16 \), \( F=-29.4 \), \( G = 16 \), \( H=-39.2 \)