Answer
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.
Solution
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.
Reviewed and approved by the UpStudy tutoring team
Explain
Simplify this solution
