Question

a projectile is launched horizontally from a cliff top at 18 m/s. determine the x - y positions at 1 - second intervals. the launch position is (0 m, 0 m). using g = 9.8 m/s/s. enter - for left and down.

Explanation:

Step1: Calculate x - Position

The horizontal motion of a projectile (assuming no air - resistance) is a uniform motion. The formula for the x - position is \(x = v_{0x}t\), where \(v_{0x}=18\ m/s\) (initial horizontal velocity) and \(t\) is the time.

• For \(t = 1.0\ s\): \(x_1=18\times1 = 18\ m\)
• For \(t = 2.0\ s\): \(x_2=18\times2=36\ m\)
• For \(t = 3.0\ s\): \(x_3=18\times3 = 54\ m\)
• For \(t = 4.0\ s\): \(x_4=18\times4=72\ m\)

Step2: Calculate y - Position

The vertical motion of a projectile is a free - fall motion. The formula for the y - position (taking downwards as negative, \(y = y_0+v_{0y}t-\frac{1}{2}gt^{2}\), and since \(y_0 = 0\ m\) and \(v_{0y}=0\ m/s\)), so \(y=-\frac{1}{2}gt^{2}\)

• For \(t = 1.0\ s\): \(y_1=-\frac{1}{2}\times9.8\times1^{2}=- 4.9\ m\)
• For \(t = 2.0\ s\): \(y_2=-\frac{1}{2}\times9.8\times2^{2}=-\frac{1}{2}\times9.8\times4=-19.6\ m\)
• For \(t = 3.0\ s\): \(y_3=-\frac{1}{2}\times9.8\times3^{2}=-\frac{1}{2}\times9.8\times9=-44.1\ m\)
• For \(t = 4.0\ s\): \(y_4=-\frac{1}{2}\times9.8\times4^{2}=-\frac{1}{2}\times9.8\times16=-78.4\ m\)

Answer:

time (s)	x - Position (m)	y - Position (m)
\(1.0\)	\(18\)	\(-4.9\)
\(2.0\)	\(36\)	\(-19.6\)
\(3.0\)	\(54\)	\(-44.1\)
\(4.0\)	\(72\)	\(-78.4\)