Question

1. an arrow is shot from a bow straight upwards at an initial speed of 104 m/s. how high in the air does the arrow fly before it comes back down?

Explanation:

Step1: Identify the kinematic equation

We use the kinematic equation \(v^{2}=v_{0}^{2}+2ah\). At the maximum height, the final velocity \(v = 0\). The acceleration \(a=-g=- 9.8\ m/s^{2}\) (negative because it acts against the motion), and the initial velocity \(v_{0}=104\ m/s\).

Step2: Rearrange the equation for height \(h\)

From \(v^{2}=v_{0}^{2}+2ah\), we can solve for \(h\). Rearranging gives \(h=\frac{v^{2}-v_{0}^{2}}{2a}\).
Substitute \(v = 0\), \(v_{0}=104\ m/s\), and \(a=-9.8\ m/s^{2}\) into the formula:

$$h=\frac{0-(104)^{2}}{2\times(-9.8)}$$

$$h=\frac{- 10816}{-19.6}$$

$$h = 551.8367\ m$$

Answer:

The arrow flies approximately \(552\ m\) high.