This problem involves projectile motion, where the archer fires an arrow straight upward with an initial velocity $v_0 = 29.1 \, \text{m/s}$. Air resistance is neglected, and we are tasked with finding the maximum height and the total time the arrow spends in the air.

Part (a): Maximum Height

To find the maximum height $h$, we use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:

$v^2 = v_0^2 + 2a \cdot h$

At maximum height, the final velocity $v$ is zero ($v = 0$). The acceleration $a$ is due to gravity, which acts downward, so $a = -g$, where $g \approx 9.8 \, \text{m/s}^2$.

Substituting these values into the equation:

$0 = v_0^2 - 2gh$

Solving for $h$:

$h = \frac{v_0^2}{2g}$

Now, substitute $v_0 = 29.1 \, \text{m/s}$ and $g = 9.8 \, \text{m/s}^2$:

$h = \frac{(29.1)^2}{2 \times 9.8}$

Let's calculate the value of $h$.

The maximum height $h$ reached by the arrow is approximately $43.2 \, \text{m}$.

Part (b): Time in the Air

The total time the arrow is in the air, $t_f$, consists of the time to reach the maximum height and the time to fall back down. Since the motion is symmetrical, the time to reach the maximum height is equal to the time to fall back down.

We can use the following kinematic equation to find the time to reach the maximum height:

$v = v_0 + at$

At maximum height, $v = 0$ and $a = -g$:

$0 = v_0 - g t_{\text{up}}$

Solving for $t_{\text{up}}$:

$t_{\text{up}} = \frac{v_0}{g}$

The total time in the air $t_f$ is twice this value:

$t_f = 2 \times t_{\text{up}} = \frac{2v_0}{g}$

Thus, the expression for the total time the arrow is in the air is:

$t_f = \frac{2v_0}{g}$

Summary:

• $h = 43.2 \, \text{m}$
• $t_f = \frac{2v_0}{g}$

Would you like more details, or do you have any further questions?

Here are five related questions to expand your understanding:

1. How would air resistance affect the maximum height of the arrow?
2. What would be the total time in the air if the initial velocity were doubled?
3. How does the acceleration due to gravity influence the time the arrow stays in the air?
4. How would the problem change if the arrow were fired at an angle instead of vertically?
5. What happens to the velocity of the arrow just before it hits the ground?

Tip: In projectile motion, the time to reach maximum height is always equal to the time to return to the launch point, assuming no air resistance.