Question

You are standing in a hole that is 6.77 m deep. You shoot an arrow with a velocity of 25.9 m/s at an angle of 50o. How far from you will the arrow hit the ground?

Answer 1

Question

Vertical speed is $v_{\perp} = v \cdot \sin \alpha$, where $v = 26.6 \frac{m}{s}$ and $\alpha = 66.2{}^{\circ}$. We need to know what is the tallest obstacle: $h$.

We will have:

$$\begin{array}{l} v_{\text{tallest}} = v \cdot \sin \alpha - g \cdot t = 0 \Rightarrow t = \frac{v \cdot \sin \alpha}{g} \Rightarrow \\ \Rightarrow h = v \cdot \sin \alpha \cdot t - \frac{g \cdot t^2}{2} = v \cdot \sin \alpha \cdot \frac{v \cdot \sin \alpha}{g} - \frac{g}{2} \cdot \frac{v^2 \cdot \sin^2 \alpha}{g^2} = \\ = \frac{v^2 \cdot \sin^2 \alpha}{g} - \frac{v^2 \cdot \sin^2 \alpha}{2g^2} = \frac{v^2 \cdot \sin^2 \alpha}{2g^2} \end{array}$$

So, according to the given values:

$$h = \frac{v^2 \cdot \sin^2 \alpha}{2g^2} = \frac{(26.6)^2 \cdot \sin^2 66.2{}^{\circ}}{2 \cdot (9.8)^2} = 3.1 \, \text{m}.$$

Answer: 3.1 meters.