Question

1. given:

initial horizontal velocity (vₓ₀): 2.39 m/s
height of elevated surface (y): 63.5 m
angle (θ): 0°
wanted: distance (x) in m?

Explanation:

Step1: Analyze vertical motion (free - fall)

In vertical direction, the initial vertical velocity \(v_{y0}=v_{x0}\sin\theta\). Since \(\theta = 0^{\circ}\), \(\sin\theta=0\), so \(v_{y0} = 0\ m/s\). The vertical displacement \(y=- 63.5\ m\) (taking downwards as negative), and the acceleration \(a = g=- 9.8\ m/s^{2}\) (taking downwards as negative). We use the equation \(y=v_{y0}t+\frac{1}{2}at^{2}\). Substituting the values, we get \(-63.5=0\times t+\frac{1}{2}\times(- 9.8)t^{2}\). Simplifying, \(-63.5=-4.9t^{2}\), then \(t^{2}=\frac{63.5}{4.9}\), and \(t=\sqrt{\frac{63.5}{4.9}}\approx\sqrt{12.96}\approx3.6\ s\).

Step2: Analyze horizontal motion (uniform motion)

In horizontal direction, there is no acceleration (\(a_x = 0\)), so the horizontal displacement \(x = v_{x0}t\). We know \(v_{x0}=2.39\ m/s\) and \(t\approx3.6\ s\). Then \(x=2.39\times3.6\approx8.60\ m\).

Answer:

The horizontal distance \(x\) is approximately \(\boldsymbol{8.60\ m}\) (the value may vary slightly depending on the precision of \(t\) calculation, for example, if we calculate \(t=\sqrt{\frac{63.5}{4.9}}\approx3.601\ s\), then \(x = 2.39\times3.601\approx8.61\ m\))