14. A hunter aims directly at a target (on the same level)38.0 m away.(a) If the arrow leaves the bow at a speed of 23.1m/s by how much will it miss the target? (b) At what angle should the bow be aimed so the target will be hit?

To solve this problem, we need to consider the horizontal and vertical components of the arrow's motion separately.<br /><br />(a) To find out by how much the arrow will miss the target, we need to calculate the horizontal distance covered by the arrow in the time it takes to reach the target.<br /><br />Given:<br />- Distance to the target (horizontal distance), \(d = 38.0 \, \text{m}\)<br />- Initial speed of the arrow, \(v = 23.1 \, \text{m/s}\)<br /><br />Since the arrow is aimed directly at the target, the horizontal component of the motion is uniform. Therefore, the time taken to reach the target is:<br /><br />\[ t = \frac{d}{v} = \frac{38.0 \, \text{m}}{23.1 \, \text{m/s}} \approx 1.64 \, \text{s} \]<br /><br />Now, we need to calculate the vertical displacement of the arrow during this time. The vertical displacement can be calculated using the formula for the vertical motion of an object under gravity:<br /><br />\[ y = v_{0y} t - \frac{1}{2} g t^2 \]<br /><br />where:<br />- \(v_{0y}\) is the initial vertical velocity of the arrow,<br />- \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)).<br /><br />Since the arrow is aimed directly at the target, the initial vertical velocity \(v_{0y}\) is 0. Therefore, the vertical displacement is:<br /><br />\[ y = -\frac{1}{2} g t^2 \]<br /><br />Substituting the values:<br /><br />\[ y = -\frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (1.64 \, \text{s})^2 \approx -12.7 \, \text{m} \]<br /><br />The negative sign indicates that the arrow falls short of the target. Therefore, the arrow will miss the target by approximately 12.7 meters.<br /><br />(b) To find the angle at which the bow should be aimed so that the arrow will hit the target, we need to ensure that the horizontal and vertical components of the motion are balanced.<br /><br />Given:<br />- Distance to the target, \(d = 38.0 \, \text{m}\)<br />- Initial speed of the arrow, \(v = 23.1 \, \text{m/s}\)<br />- Acceleration due to gravity, \(g = 9.8 \, \text{m/s}^2\)<br /><br />We need to find the angle \(\theta\) such that the vertical displacement is zero. This can be done by using the equation for the range of a projectile:<br /><br />\[ R = \frac{v^2 \sin(2\theta)}{g} \]<br /><br />where \(R\) is the range (horizontal distance covered).<br /><br />We want the range to be equal to the distance to the target:<br /><br />\[ 38.0 \, \text{m} = \frac{23.1^2 \sin(2\theta)}{9.8} \]<br /><br />Solving for \(\sin(2\theta)\):<br /><br />\[ \sin(2\theta) = \frac{38.0 \times 9.8}{23.1^2} \approx 0.617 \]<br /><br />Now, find the angle \(\theta\):<br /><br />\[ 2\theta = \sin^{-1}(0.617) \]<br /><br />\[ 2\theta \approx 38.7^\circ \]<br /><br />\[ \theta \approx 19.4^\circ \]<br /><br />Therefore, the bow should be aimed at an angle of approximately 19.4 degrees above the horizontal to hit the target.