2 A ball is thrown vertically upwards. Its height, , above the ground at time seconds after being
thrown is given by the formula .
(a) State the height above the ground from which the ball is thrown.
(b) Express in the form where and are constants to be determined.
© Hence state the maximum height attained by the ball and the time at which this occurs.
(l)

Ask by Graham Mccarthy. in Singapore

Jan 12,2025

Question

2 A ball is thrown vertically upwards. Its height, , above the ground at time seconds after being
thrown is given by the formula .
(a) State the height above the ground from which the ball is thrown.
(b) Express in the form where and are constants to be determined.
© Hence state the maximum height attained by the ball and the time at which this occurs.
(l)

Ask by Graham Mccarthy. in Singapore

Jan 12,2025

UpStudy AI · Accepted Answer

Let's analyze the motion of the ball thrown vertically upwards using the given height function:

$$
h(t) = 1.75 + 5t - 5t^2
$$

where:
- $$ h(t) $$ is the height in meters,
- $$ t $$ is the time in seconds.

### **Part (a): Initial Height Above the Ground**

**Question:**  
_State the height above the ground from which the ball is thrown._

**Solution:**  
The initial height is the height of the ball at time $$ t = 0 $$.

$$
h(0) = 1.75 + 5(0) - 5(0)^2 = 1.75 \, \text{meters}
$$

**Answer:**  
The ball is thrown from a height of **1.75 meters** above the ground.

---

### **Part (b): Expressing $$ h $$ in the Form $$ a + b(t + c)^2 $$**

**Question:**  
_Express $$ h $$ in the form $$ a + b(t + c)^2 $$ where $$ a $$, $$ b $$, and $$ c $$ are constants._

**Solution:**  
We start with the original equation:

$$
h(t) = 1.75 + 5t - 5t^2
$$

To express this in the form $$ a + b(t + c)^2 $$, we'll complete the square.

1. **Factor out the coefficient of $$ t^2 $$ from the first two terms:**

$$
h(t) = -5t^2 + 5t + 1.75 = -5\left(t^2 - t\right) + 1.75
$$

2. **Complete the square inside the parentheses:**

$$
t^2 - t = \left(t - \frac{1}{2}\right)^2 - \frac{1}{4}
$$

3. **Substitute back into the equation:**

$$
h(t) = -5\left[\left(t - \frac{1}{2}\right)^2 - \frac{1}{4}\right] + 1.75 = -5\left(t - \frac{1}{2}\right)^2 + \frac{5}{4} + 1.75
$$

4. **Simplify the constants:**

$$
\frac{5}{4} + 1.75 = 1.25 + 1.75 = 3.0
$$

5. **Rewrite in the desired form:**

$$
h(t) = 3.0 - 5\left(t - \frac{1}{2}\right)^2 = 3.0 + (-5)\left(t + \left(-\frac{1}{2}\right)\right)^2
$$

Comparing this to $$ a + b(t + c)^2 $$:

- $$ a = 3.0 $$
- $$ b = -5 $$
- $$ c = -0.5 $$

**Answer:**  
After completing the square, the height can be written as

$$
h = 3.0 - 5\left(t - 0.5\right)^2
$$

where:
- $$ a = 3.0 $$,
- $$ b = -5 $$,
- $$ c = -0.5 $$.

---

### **Part (c): Maximum Height and Time of Occurrence**

**Question:**  
_Hence state the maximum height attained by the ball and the time at which this occurs._

**Solution:**  
From the expression obtained in part (b):

$$
h(t) = 3.0 - 5\left(t - 0.5\right)^2
$$

This is a downward-opening parabola (since the coefficient of the squared term is negative). The maximum height occurs at the vertex of the parabola.

- **Time of Maximum Height ($$ t_{max} $$)**:  
  The vertex form $$ a + b(t + c)^2 $$ reaches its maximum when $$ t = -c $$.

$$
  c = -0.5 \quad \Rightarrow \quad t_{max} = -(-0.5) = 0.5 \, \text{seconds}
  $$

- **Maximum Height ($$ h_{max} $$)**:  
  Substitute $$ t = 0.5 $$ into the height equation:

$$
  h(0.5) = 3.0 - 5(0.5 - 0.5)^2 = 3.0 - 5(0)^2 = 3.0 \, \text{meters}
  $$

**Answer:**  
The ball reaches a **maximum height of 3.0 meters** at **0.5 seconds** after being thrown.
The ball is thrown from a height of 1.75 meters. Expressed in the form $$ a + b(t + c)^2 $$, the height is $$ 3.0 - 5(t - 0.5)^2 $$. The maximum height is 3.0 meters, occurring at 0.5 seconds.

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