16. an arrow is shot from a composite bow horizontally towards a target 105 m away. if the arrow leaves the…

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.