To solve this problem, we can use the kinematic equations for motion under constant acceleration due to gravity. Here's the plan:

1. Initial velocity of the bullet: The total time the bullet is in the air is 60 seconds. Since the bullet goes up and then comes back down, the time to reach its maximum height is half the total time, i.e., $t_{\text{up}} = \frac{60}{2} = 30 \, \text{seconds}$.

The formula for the final velocity when an object reaches its maximum height (where its velocity is zero) is:

$v_f = v_0 + a t$

Where:

• $v_f = 0 \, \text{m/s}$ (velocity at maximum height),
• $v_0$ is the initial velocity,
• $a = -9.8 \, \text{m/s}^2$ (acceleration due to gravity),
• $t = 30 \, \text{s}$ (time to reach maximum height).

Substitute the values:

$0 = v_0 - 9.8 \times 30$

Solving for $v_0$:

$v_0 = 9.8 \times 30 = 294 \, \text{m/s}$

So, the initial velocity of the bullet is 294 m/s.

2. Maximum height reached by the bullet: To find the maximum height, we use the equation:

$h_{\text{max}} = v_0 t + \frac{1}{2} a t^2$

Substitute the known values:

$h_{\text{max}} = 294 \times 30 + \frac{1}{2} \times (-9.8) \times (30)^2$

Simplifying:

$h_{\text{max}} = 8820 - 4410 = 4410 \, \text{m}$

Thus, the maximum height reached by the bullet is 4410 meters.

Summary:

• The initial velocity of the bullet is 294 m/s.
• The maximum height reached is 4410 meters.

Would you like further clarification or details on any step?

Here are some related questions:

1. What would happen if the acceleration due to gravity were different?
2. How does air resistance affect the maximum height of the bullet?
3. What would the time of flight be if the bullet was fired at an angle?
4. How would you calculate the speed of the bullet after 15 seconds in the air?
5. How would the results change if the initial velocity were halved?

Tip: Always double-check units and signs when dealing with kinematic equations, especially for acceleration due to gravity, which is negative for upward motion.