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

The initial horizontal velocity \(v_{0x}\) is given as \(178\ km/h\). To convert it to \(m/s\), we use the conversion factor \(1\ km = 1000\ m\) and \(1\ h=3600\ s\).
\(v_{0x}=178\times\frac{1000}{3600}\ m/s=\frac{1780}{36}\ m/s\approx 49.44\ m/s\)

Step2: Find the time of flight

In the horizontal direction (\(x - \)direction), the motion is uniform (\(a_x = 0\)). The horizontal displacement \(x = v_{0x}t\). We know \(x = 105\ m\) and \(v_{0x}\approx49.44\ m/s\). Solving for \(t\):
\(t=\frac{x}{v_{0x}}=\frac{105}{49.44}\ s\approx 2.12\ s\)

Step3: Find the vertical - drop

In the vertical direction (\(y - \)direction), the initial vertical velocity \(v_{0y}=0\ m/s\) and the acceleration \(a = g= 9.8\ m/s^{2}\). Using the equation \(y=v_{0y}t+\frac{1}{2}at^{2}\), since \(v_{0y} = 0\), we have \(y=\frac{1}{2}gt^{2}\)
Substitute \(t = 2.12\ s\) and \(g = 9.8\ m/s^{2}\)
\(y=\frac{1}{2}\times9.8\times(2.12)^{2}\)
\(y = 4.9\times4.4944\)
\(y\approx22.0\ m\)

Answer:

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