We can solve this problem using the basic equations of motion under constant acceleration due to gravity.

1. Maximum Height $h$:

The formula to find the maximum height when an object is launched vertically with an initial velocity $v_0$ is:

$v^2 = v_0^2 - 2gh$

At the maximum height, the final velocity $v = 0$. So, solving for $h$:

$0 = v_0^2 - 2gh$

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

Where:

• $v_0 = 29.1 \, \text{m/s}$ is the initial velocity,
• $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.

Substitute the known values:

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

Let's calculate the height:

$h = \frac{847.81}{19.6} = 43.26 \, \text{m}$

So, the maximum height reached by the arrow is 43.26 meters.

2. Time of Flight:

To find the time the arrow is in the air, we can use the formula for the time it takes to reach maximum height:

$v = v_0 - gt$

At maximum height, $v = 0$, so:

$0 = v_0 - gt$

$t = \frac{v_0}{g}$

This is the time to reach the maximum height. To find the total time of flight, we double this value because the time taken to go up is the same as the time taken to come down:

$T = 2 \times \frac{v_0}{g}$

Substitute the known values:

$T = 2 \times \frac{29.1}{9.8} = 5.94 \, \text{seconds}$

Thus, the total time the arrow is in the air is 5.94 seconds.

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Would you like further details on any part of the solution? Here are some related questions to explore:

1. How does air resistance affect the maximum height of the arrow?
2. What would happen to the maximum height if the initial velocity were doubled?
3. How would the time of flight change if gravity on another planet is different?
4. Can we apply energy conservation to solve for the maximum height?
5. What is the velocity of the arrow just before it hits the ground?

Tip: When solving projectile motion problems, remember that horizontal and vertical motions are independent.