Question

If Jack tosses the phone to Darnell with an initial velocity of 35 feet per second, an equation for the height h in feet of the phone after t seconds is h =-16t^2 + 35t +5.

Answer 1

The equation given for the height \( h \) of the phone after \( t \) seconds is:

\[
h = -16t^2 + 35t + 5
\]

This is a quadratic equation in the standard form \( h = at^2 + bt + c \) where:
- \( a = -16 \),
- \( b = 35 \),
- \( c = 5 \).

### Key Characteristics of the Equation:

1. **Direction of Motion**: Since the coefficient of \( t^2 \) (which is \( a = -16 \)) is negative, the parabola opens downward, indicating that the phone will reach a maximum height and then fall back down.

2. **Finding the Maximum Height**:
The time at which the maximum height is reached can be found using the vertex formula for a parabola given by \( t = -\frac{b}{2a} \).

Substituting the values of \( b \) and \( a \):

\[
t = -\frac{35}{2(-16)} = \frac{35}{32} \approx 1.094 \text{ seconds}
\]

Now, substituting \( t \) back into the height equation to find the maximum height \( h \):

\[
h = -16\left(\frac{35}{32}\right)^2 + 35\left(\frac{35}{32}\right) + 5
\]

Calculating \( \left(\frac{35}{32}\right)^2 = \frac{1225}{1024} \):

\[
h = -16\left(\frac{1225}{1024}\right) + \frac{1225}{32} + 5
\]

Now calculating each term:

\[
-16\left(\frac{1225}{1024}\right) = -\frac{19600}{1024}
\]
\[
35\left(\frac{35}{32}\right) = \frac{1225}{32} = \frac{1225 \times 32}{1024} = \frac{39200}{1024}
\]
\[
5 = \frac{5 \times 1024}{1024} = \frac{5120}{1024}
\]

Combining these:

\[
h = -\frac{19600}{1024} + \frac{39200}{1024} + \frac{5120}{1024} = \frac{39200 - 19600 + 5120}{1024} = \frac{24520}{1024} \approx 23.95 \text{ feet}
\]

So the maximum height reached by the phone is approximately **23.95 feet**.

3. **Finding the Time When the Phone Hits the Ground**:
To find when the phone hits the ground, set \( h = 0 \):

\[
-16t^2 + 35t + 5 = 0
\]

You can apply the quadratic formula:

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

Substituting \( a = -16 \), \( b = 35 \), and \( c = 5 \):

\[
t = \frac{-35 \pm \sqrt{35^2 - 4 \cdot (-16) \cdot 5}}{2 \cdot (-16)}
\]

Calculate the discriminant:

\[
35^2 = 1225
\]
\[
4 \cdot 16 \cdot 5 = 320
\]
\[
\Rightarrow 1225 + 320 = 1545
\]

Now, find \( t \):

\[
t = \frac{-35 \pm \sqrt{1545}}{-32}
\]

Approximating \( \sqrt{1545} \approx 39.3 \):

\[
t = \frac{-35 + 39.3}{-32} \quad \text{and} \quad t = \frac{-35 - 39.3}{-32}
\]

Calculating both:

\[
t_1 = \frac{4.3}{-32} \quad \text{(not valid, as time cannot be negative)}
\]
\[
t_2 = \frac{-74.3}{-32} \approx 2.32 \text{ seconds}
\]

Thus, the phone will hit the ground approximately at **2.32 seconds**.