Question

1. an arrow is shot from a composite bow horizontally towards a target 105 m away. if the arrow leaves the bow at 178 km/h, how far does it drop before it reaches the length of the target?

Explanation:

Step1: Convert the initial velocity to m/s

We know that \(1\ km = 1000\ m\) and \(1\ h=3600\ s\).
The initial horizontal velocity \(v_{0x}\):
\(v_{0x}=178\ km/h=\frac{178\times1000}{3600}\ m/s\approx49.44\ m/s\)

Step2: Find the time of flight

In the horizontal direction (\(a_x = 0\)), the horizontal displacement \(x = v_{0x}t\).
We want to find \(t\) when \(x = 105\ m\).
\(t=\frac{x}{v_{0x}}=\frac{105}{49.44}\ s\approx2.12\ s\)

Step3: Find the vertical - drop

In the vertical direction (\(v_{0y} = 0\), \(a_y=g = 9.8\ m/s^{2}\)), the vertical displacement \(y\) is given by the equation \(y=v_{0y}t+\frac{1}{2}a_yt^{2}\).
Since \(v_{0y} = 0\), \(y=\frac{1}{2}gt^{2}\)
Substitute \(t = 2.12\ s\) and \(g = 9.8\ m/s^{2}\) into the formula:
\(y=\frac{1}{2}\times9.8\times(2.12)^{2}\)
\(y = 4.9\times4.4944\)
\(y\approx22.02\ m\)

Answer:

The arrow drops approximately \(22.0\ m\) before it reaches the target.