Question

1. given:

initial horizontal velocity (vₓ₀): 3.08 m/s
height of elevated surface (y): 8.14 m
angle (θ): 0°
wanted: distance (x) in m?
distance? m

Explanation:

Step1: Find time of fall using vertical motion

The vertical motion is free - fall with initial vertical velocity \(v_{y0}=v_0\sin\theta\). Since \(\theta = 0^{\circ}\), \(v_{y0}=0\ m/s\). The vertical displacement \(y = v_{y0}t+\frac{1}{2}gt^{2}\), substituting \(v_{y0} = 0\), we get \(y=\frac{1}{2}gt^{2}\). Solving for \(t\): \(t=\sqrt{\frac{2y}{g}}\), where \(g = 9.8\ m/s^{2}\) and \(y = 8.14\ m\).
\(t=\sqrt{\frac{2\times8.14}{9.8}}=\sqrt{\frac{16.28}{9.8}}\approx\sqrt{1.6612}\approx1.289\ s\)

Step2: Find horizontal distance using horizontal motion

In horizontal motion (no air resistance), the horizontal velocity \(v_{x}=v_{x0}\) (constant). The horizontal distance \(x = v_{x0}t\), where \(v_{x0}=3.08\ m/s\) and \(t\approx1.289\ s\).
\(x=3.08\times1.289\approx3.08\times1.29\approx3.97\ m\) (more accurately, \(3.08\times1.289 = 3.08\times(1 + 0.2+0.08 + 0.009)=3.08+0.616+0.2464 + 0.02772=3.97012\ m\))

Answer:

\(\approx3.97\)