To determine the maximum height of the arrow, we can use the following physics concepts of projectile motion.
-
Break down the initial velocity into its components: The initial velocity \(v_0 = 63 , \text{m/s}\) at an angle of \(45^\circ\) can be resolved into horizontal and vertical components: \[ v_{0x} = v_0 \cos(45^\circ) = 63 \cos(45^\circ) = 63 \times \frac{\sqrt{2}}{2} \approx 44.5 , \text{m/s} \] \[ v_{0y} = v_0 \sin(45^\circ) = 63 \sin(45^\circ) = 63 \times \frac{\sqrt{2}}{2} \approx 44.5 , \text{m/s} \]
-
Use the vertical motion equations: We will use the kinematic equation for vertical motion to find the maximum height \(H\). At maximum height, the final vertical velocity \(v_y = 0 , \text{m/s}\). The relevant equation is: \[ v_y^2 = v_{0y}^2 - 2gH \] where \(g \approx 9.81 , \text{m/s}^2\) is the acceleration due to gravity.
Plugging in \(v_y = 0\): \[ 0 = (44.5)^2 - 2 \cdot 9.81 \cdot H \] \[ (44.5)^2 = 2 \cdot 9.81 \cdot H \] \[ 1980.25 = 19.62 \cdot H \] \[ H = \frac{1980.25}{19.62} \approx 101.86 , \text{m} \]
Therefore, the maximum height of the arrow is approximately 101.86 meters.
2 answers