Question

1. given: initial horizontal velocity (vₓ₀): 88.1 m/s height of elevated surface (y): 200 m angle (θ): 0° wanted: distance (x) in m?

Explanation:

Step1: Find time of fall using vertical motion

The vertical motion is free - fall with initial vertical velocity \(v_{y0}=v_{x0}\sin\theta\). Since \(\theta = 0^{\circ}\), \(v_{y0}=0\). The vertical displacement formula is \(y = v_{y0}t+\frac{1}{2}gt^{2}\). Substituting \(v_{y0} = 0\), we get \(y=\frac{1}{2}gt^{2}\). Solving for \(t\): \(t=\sqrt{\frac{2y}{g}}\), where \(y = 200\space m\) and \(g = 9.8\space m/s^{2}\).
\(t=\sqrt{\frac{2\times200}{9.8}}=\sqrt{\frac{400}{9.8}}\approx\sqrt{40.816}\approx6.39\space s\)

Step2: Find horizontal distance using horizontal motion

Horizontal motion is uniform motion with \(v_{x}=v_{x0}\) (since no air resistance). The formula for horizontal distance \(x = v_{x0}t\). Substituting \(v_{x0}=88.1\space m/s\) and \(t\approx6.39\space s\).
\(x = 88.1\times6.39\approx88.1\times6.39\approx563\space m\) (more precisely, \(t=\sqrt{\frac{400}{9.8}}\approx6.3887\space s\), \(x = 88.1\times6.3887\approx88.1\times6.3887\approx562.84\space m\))

Answer:

\(\approx563\) (or more precisely \(\approx562.8\))