Question

a projectile is launched horizontally from a cliff top at 15 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: Horizontal velocity component

In projectile motion, the horizontal velocity \(v_x\) remains constant (assuming no air - resistance). The initial horizontal velocity \(v_{x0}=15\ m/s\). So for all times \(t\), \(v_x = 15\ m/s\).

• For \(t = 1.0\ s\), \(v_x(A)=15\ m/s\)
• For \(t = 2.0\ s\), \(v_x(C)=15\ m/s\)
• For \(t = 3.0\ s\), \(v_x(E)=15\ m/s\)
• For \(t = 4.0\ s\), \(v_x(G)=15\ m/s\)

Step2: Vertical velocity component

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

• When \(t = 1.0\ s\), \(v_y(B)=v_{y0}+gt=0+( - 9.8)\times1=-9.8\ m/s\)
• When \(t = 2.0\ s\), \(v_y(D)=v_{y0}+gt=0+( - 9.8)\times2=-19.6\ m/s\)
• When \(t = 3.0\ s\), \(v_y(F)=v_{y0}+gt=0+( - 9.8)\times3=-29.4\ m/s\)
• When \(t = 4.0\ s\), \(v_y(H)=v_{y0}+gt=0+( - 9.8)\times4=-39.2\ m/s\)

Answer:

time (s)	\(V_x(m/s)\)	\(V_y(m/s)\)
\(1.0\)	\(15\)	\(-9.8\)
\(2.0\)	\(15\)	\(-19.6\)
\(3.0\)	\(15\)	\(-29.4\)
\(4.0\)	\(15\)	\(-39.2\)