Use the equation for the vertical motion:

−
16
𝑡
2
+
45.72
𝑡
+
2.5
=
10
−16t 
2
 +45.72t+2.5=10
Solve this quadratic equation for 
𝑡
t.

Then, plug the values of 
𝑡
t into the horizontal motion equation:

𝑥
(
𝑡
)
=
168.39
𝑡
x(t)=168.39t

Let's start by solving the vertical motion equation first:

\[
-16t^2 + 45.72t + 2.5 = 10
\]

### Step 1: Move all terms to one side of the equation
Subtract 10 from both sides to set the equation equal to zero:

\[
-16t^2 + 45.72t + 2.5 - 10 = 0
\]

Simplifying the constants:

\[
-16t^2 + 45.72t - 7.5 = 0
\]

This is a standard quadratic equation in the form \( at^2 + bt + c = 0 \), where \( a = -16 \), \( b = 45.72 \), and \( c = -7.5 \).

### Step 2: Use the quadratic formula
The quadratic formula is:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substitute \( a = -16 \), \( b = 45.72 \), and \( c = -7.5 \):

\[
t = \frac{-45.72 \pm \sqrt{(45.72)^2 - 4(-16)(-7.5)}}{2(-16)}
\]

### Step 3: Solve the discriminant
First, calculate the discriminant:

\[
(45.72)^2 = 2089.7184
\]
\[
4(-16)(-7.5) = 480
\]
\[
\text{Discriminant} = 2089.7184 - 480 = 1609.7184
\]

### Step 4: Solve for \( t \)
Now, we can calculate \( t \):

\[
t = \frac{-45.72 \pm \sqrt{1609.7184}}{-32}
\]
\[
\sqrt{1609.7184} \approx 40.12
\]
\[
t = \frac{-45.72 \pm 40.12}{-32}
\]

We have two possible solutions for \( t \):

\[
t_1 = \frac{-45.72 + 40.12}{-32} = \frac{-5.6}{-32} \approx 0.175
\]
\[
t_2 = \frac{-45.72 - 40.12}{-32} = \frac{-85.84}{-32} \approx 2.68
\]

### Step 5: Plug the values of \( t \) into the horizontal motion equation

The horizontal motion equation is given by:

\[
x(t) = 168.39t
\]

For \( t_1 \approx 0.175 \):

\[
x(0.175) = 168.39 \times 0.175 \approx 29.47
\]

For \( t_2 \approx 2.68 \):

\[
x(2.68) = 168.39 \times 2.68 \approx 451.31
\]

### Final Answer:
- \( t_1 \approx 0.175 \), so \( x(0.175) \approx 29.47 \)
- \( t_2 \approx 2.68 \), so \( x(2.68) \approx 451.31 \)

These are the solutions for the time \( t \) and corresponding horizontal distance \( x(t) \).

Do you want more details or have any questions?

Here are 5 related questions to explore:
1. How does the initial height affect the vertical motion equation?
2. What would happen to the motion if gravity were different?
3. How would you solve this equation graphically?
4. What if the initial velocity was changed in the vertical equation?
5. How do we interpret the meaning of both \( t_1 \) and \( t_2 \) physically?

**Tip**: Always check if both values of \( t \) make sense in the physical context (e.g., time cannot be negative).

Use the equation for the vertical motion: −16t^2 + 45.72t + 2.5 = 10. Solve this quadratic equation for t. Then, plug the values of t into the horizontal motion equation: x(t) = 168.39t.

Learn how to solve the quadratic equation −16t^2 + 45.72t + 2.5 = 10 for time, then use it to calculate horizontal distance using x(t) = 168.39t. This problem integrates algebraic and physics concepts, including the quadratic formula and projectile motion equations. Suitable for Grades 10-12.

Math Problem Statement

− 16 𝑡 2 + 45.72 𝑡 + 2.5

𝑥 ( 𝑡 )

Solution

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