Question

An arrow is shot with an initial velocity of 63. m/s at an angle of 45° above the horizontal. What is the maximum height of the arrow?

Answer 1

To determine the maximum height of the arrow, we can use the following physics concepts of projectile motion.

1. **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:
\[
v_{0x} = v_0 \cos(45^\circ) = 63 \cos(45^\circ) = 63 \times \frac{\sqrt{2}}{2} \approx 44.5 \, \text{m/s}
\]
\[
v_{0y} = v_0 \sin(45^\circ) = 63 \sin(45^\circ) = 63 \times \frac{\sqrt{2}}{2} \approx 44.5 \, \text{m/s}
\]

2. **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**.