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Let's go through the questions one by one and solve them using standard kinematic equations for vertical motion. We'll assume upward is the positive direction and gravity acts downward with acceleration \( g \).

The initial velocity is given as \( u_0 \), and we need to express our answers in terms of \( g \) and \( u_0 \).

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**(i) How long does it take to reach its maximum height?**

At maximum height, the velocity becomes zero.

From the equation of motion:
\[
v = u_0 - g t
\]
At maximum height, \( v = 0 \), so:
\[
0 = u_0 - g t_{\text{max}}
\]
Thus, the time to reach maximum height is:
\[
t_{\text{max}} = \frac{u_0}{g}
\]

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**(ii) How long does it take to return to its starting point?**

The total time for the stone to go up and come back down is twice the time it takes to reach the maximum height (because the motion is symmetric):
\[
t_{\text{total}} = 2 \times t_{\text{max}} = 2 \times \frac{u_0}{g} = \frac{2u_0}{g}
\]

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**(iii) How fast should a second stone, thrown vertically upwards at the same time as the first reaches its maximum height, be traveling in order to collide with the first stone when at the second stone's maximum height?**

At the moment the second stone is thrown, the first stone is at its maximum height and begins to fall. Let's call the speed of the second stone \( u_1 \).

For the two stones to collide at the second stone's maximum height:
- The time for the second stone to reach its maximum height must equal the time for the first stone to fall from its maximum height to the same point.
- The time for the second stone to reach its maximum height is \( t_1 = \frac{u_1}{g} \).
- The time for the first stone to fall from maximum height is \( t_1 = t_{\text{max}} = \frac{u_0}{g} \).

Thus, the two times are equal, so:
\[
\frac{u_1}{g} = \frac{u_0}{g}
\]
This gives:
\[
u_1 = u_0
\]

Therefore, the second stone must be thrown upwards with the same speed \( u_1 = u_0 \).

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**(iv) If those activities were repeated on the moon (where \( g = 2 \, \text{m/s}^2 \) in place of the usual 10), which of your answers would be unchanged, if any?**

The equations for the time to reach maximum height, total time, and the required speed of the second stone all involve \( g \). Since the value of \( g \) changes on the moon, the times will increase because \( g \) is smaller.

However, the ratio of the speeds required for the two stones to collide remains the same. Hence, the *form* of the answer for (iii) remains unchanged, but the times will be different due to the change in gravity.

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Do you want further details on any of these parts? Here are five additional questions for practice:

1. What would the maximum height of the stone be in terms of \( u_0 \) and \( g \)?
2. If the stone were thrown with double the initial speed \( 2u_0 \), how would the time to reach the maximum height change?
3. How would the total time change if the stone were thrown on a planet where gravity is three times Earth's gravity?
4. What would be the impact of air resistance on the time to reach the maximum height?
5. If the stone was thrown downward with speed \( u_0 \), how long would it take to reach the ground?

**Tip:** When solving problems involving vertical motion under gravity, always check for symmetry in the upward and downward motion, which simplifies calculations!

A stone is thrown vertically upwards into the air at u0 m/s. In all parts of this question, give your answer in terms of g or u0 or both. (i) How long does it take to reach its maximum height? (ii) How long does it take to return to its starting point? (iii) How fast should a second stone, thrown vertically upwards at the same time as the first reaches its maximum height, be traveling in order to collide with the first stone when at the second stone's maximum height? (iv) If those activities were repeated on the moon (where g = 2 instead of the usual 10) which of your answers would be unchanged, if any?

This problem explores vertical motion by calculating the time to reach maximum height, total return time, and the velocity of a second stone to collide mid-air. The analysis uses kinematic equations and applies changes for different gravitational fields, such as on the Moon.

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