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This problem is about a ball being thrown upwards with an initial velocity of \( v_0 = 27.0 \, \text{m/s} \). The height \( h \) of the ball as a function of time \( t \) is given by the equation:

\[
h = v_0 t - \frac{1}{2} g t^2
\]
where:
- \( v_0 = 27.0 \, \text{m/s} \) is the initial velocity,
- \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity,
- \( t \) is the time,
- \( h \) is the height of the ball above the point it was thrown from.

The two questions ask:
a) When is the ball at a height of \( 10.3 \, \text{m} \)?
b) When does the ball return to the height it was thrown from (i.e., when does \( h = 0 \))?

### Part (a): Finding when the ball reaches a height of \( 10.3 \, \text{m} \)
We use the equation \( h = 10.3 \, \text{m} \):

\[
10.3 = 27.0 t - \frac{1}{2} (9.81) t^2
\]

This is a quadratic equation in \( t \), which can be solved using the quadratic formula.

### Part (b): Finding when the ball returns to the starting height \( h = 0 \)
In this case, we set \( h = 0 \):

\[
0 = 27.0 t - \frac{1}{2} (9.81) t^2
\]

This equation can also be solved using the quadratic formula to find the time when the ball returns to the original height.

Let me calculate the exact times for both parts.### Results:

#### Part (a)
The ball reaches a height of \( 10.3 \, \text{m} \) at two different times:
- \( t_1 = 0.41 \, \text{seconds} \)
- \( t_2 = 5.09 \, \text{seconds} \)

#### Part (b)
The ball returns to the height it was thrown from (\( h = 0 \)) after:
- \( t = 5.50 \, \text{seconds} \)

Would you like more details on the steps used to solve this problem, or do you have any further questions?

### Related Questions:
1. What is the maximum height the ball reaches?
2. How long does it take to reach the maximum height?
3. What is the velocity of the ball when it reaches a height of 10.3 m?
4. How would the time change if the initial velocity was different?
5. How can you find the total distance traveled by the ball?

#### Tip:
Quadratic equations often have two solutions, so always check both to see if they apply to your problem context.

A ball is thrown straight up with an initial velocity of 27.0 m/s. Using the equation h = v_0 * t - 1/2 * g * t^2 (where g = 9.81 m/s^2), find: 
 a) When the ball reaches a height of 10.3 meters? 
 b) When the ball returns to the height it was thrown from?

This problem demonstrates how to solve a quadratic equation to find the time when a ball reaches a height of 10.3 meters, and when it returns to its original height after being thrown upwards with an initial velocity of 27.0 m/s. Suitable for high school students studying kinematics and quadratic equations.

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