Let's break down the problem into parts:

1. Height of the Coin After 0.1 Seconds:

Given:

• Initial velocity, $v_0 = 5 \, \text{m/s}$ (upward)
• Initial height, $h_0 = 1.57 \, \text{m}$
• Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$ (downward)
• Time, $t = 0.1 \, \text{seconds}$

The height of the coin at any time $t$ can be calculated using the equation:

$h(t) = h_0 + v_0 t - \frac{1}{2} g t^2$

Plugging in the values:

$h(0.1) = 1.57 + 5(0.1) - \frac{1}{2} \times 9.8 \times (0.1)^2$

$h(0.1) = 1.57 + 0.5 - 0.049$

$h(0.1) = 2.021 \, \text{m}$

2. Maximum Height Reached by the Coin:

At the maximum height, the velocity of the coin will be zero. We can use the following equation to find the maximum height $h_{\text{max}}$:

$v^2 = v_0^2 - 2g(h_{\text{max}} - h_0)$

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

$0 = 5^2 - 2 \times 9.8 \times (h_{\text{max}} - 1.57)$

$25 = 19.6 \times (h_{\text{max}} - 1.57)$

$h_{\text{max}} - 1.57 = \frac{25}{19.6}$

$h_{\text{max}} - 1.57 = 1.276$

$h_{\text{max}} = 2.846 \, \text{m}$

3. Total Time in the Air:

The time to reach the maximum height can be calculated using:

$v = v_0 - gt$

Set $v = 0$:

$0 = 5 - 9.8t_{\text{up}}$

$t_{\text{up}} = \frac{5}{9.8} \approx 0.51 \, \text{seconds}$

The total time to go up and come down will be twice this time, but we need to consider the return to the initial height:

The coin takes $0.51$ seconds to reach the maximum height, and the same amount of time to return to the original height of 1.57 meters. So, the time for the upward and downward journey is:

$t_{\text{total}} = 2 \times 0.51 = 1.02 \, \text{seconds}$

Thus, the coin will be in the air for about 1.02 seconds.

Summary:

1. Height after 0.1 seconds: 2.021 meters.
2. Maximum height reached: 2.846 meters.
3. Total time in the air: 1.02 seconds.

---

Would you like further details or explanations on any of these calculations?

Here are some related questions:

1. How would air resistance affect the maximum height?
2. What is the significance of the initial velocity in determining the maximum height?
3. How would the time in the air change if the initial height was different?
4. Can we calculate the velocity of the coin at any specific height?
5. How does changing the initial velocity affect the overall trajectory?

Tip: When solving projectile motion problems, always start by identifying and listing the known variables, and choose the appropriate kinematic equations accordingly.