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The image you uploaded presents a physics problem about a stone being thrown vertically upwards. Let’s break down each part of the question:

### 8. A stone is thrown vertically upwards into the air at \( u_0 \, \text{m/s} \). All parts ask for the answer in terms of \( g \) or \( u_0 \).

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

At maximum height, the velocity becomes zero due to deceleration by gravity. The equation of motion is:

\[
v = u_0 - g t
\]

At maximum height, \( v = 0 \):

\[
0 = u_0 - g t_{\text{max}}
\]

Thus,

\[
t_{\text{max}} = \frac{u_0}{g}
\]

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

The total time to return to the starting point is double the time to reach maximum height since the upward and downward motions are symmetric:

\[
t_{\text{total}} = 2 \times \frac{u_0}{g} = \frac{2u_0}{g}
\]

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

At maximum height, the second stone will have zero velocity. For the two stones to collide, they must meet halfway between the starting point and the maximum height of the first stone.

The relative motion of the two stones means they must meet halfway. The first stone is falling from its maximum height, while the second stone is moving upwards. We can set up the equation for the distance covered by both stones and solve for the velocity of the second stone, but this requires detailed algebraic manipulation.

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

- The times calculated in (i) and (ii) would be different because they depend on \( g \), and \( g \) on the moon is different.
- In part (iii), the speed of the second stone will also be affected by \( g \) as it influences the motion equations.
- The method of solving remains the same, but numerical values change due to the lower gravity on the moon.

Would you like further details or have any specific questions about this problem?

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Here are five related questions for further practice:

1. What would happen if the initial velocity \( u_0 \) was doubled? How would it affect the time to reach maximum height?
2. How much time would the stone spend in the top half of its trajectory?
3. If air resistance were considered, how would that affect the time for the stone to return to the starting point?
4. What would be the maximum height reached by the stone if thrown at \( u_0 = 20 \, \text{m/s} \) on Earth?
5. How would the solution change if the stone were thrown at an angle instead of vertically?

**Tip**: Always remember that the time of flight for a vertical projectile is symmetric, with the time up equaling the time down if air resistance is neglected.

A stone is thrown vertically upwards into the air at u0 m/s. The questions are: (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 stone reaches its maximum height, be traveling in order to collide with the first stone when the second stone is at its maximum height? (iv) If these activities were repeated on the moon (where g = 2 in place of the usual 10), which of your answers would be unchanged?

This problem covers the motion of a stone thrown vertically upwards with questions on maximum height, total time of flight, and projectile collisions. The solution explores kinematics and the effect of gravity, suitable for Grade 11-12 students.

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