Problem Statement:
A person throws a ball from a height of 6 feet with an initial vertical velocity of 48 feet per second. The vertical motion model is given by the equation:
\[
h = -16t^2 + vt + s
\]
where:
- \( h \) is the height in feet,
- \( t \) is the time in seconds,
- \( v \) is the initial velocity in feet per second,
- \( s \) is the initial height in feet.
We need to calculate the amount of time the ball is in the air before it hits the ground again. When the ball hits the ground, its height \( h \) will be 0. Therefore, we need to solve for \( t \) when \( h = 0 \).
Step-by-Step Solution:
1.
Substitute the given values into the equation:
- Initial velocity \( v = 48 \) feet per second,
- Initial height \( s = 6 \) feet.
The equation becomes:
\[
h = -16t^2 + 48t + 6
\]
2.
Set \( h = 0 \) to find when the ball hits the ground:
\[
0 = -16t^2 + 48t + 6
\]
3.
Rearrange the equation into standard quadratic form:
\[
-16t^2 + 48t + 6 = 0
\]
To simplify, we can multiply the entire equation by \(-1\) (optional but makes the leading coefficient positive):
\[
16t^2 - 48t - 6 = 0
\]
4.
Identify the coefficients \( a \), \( b \), and \( c \) for the quadratic formula:
The quadratic formula is:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
From the equation \( 16t^2 - 48t - 6 = 0 \):
- \( a = 16 \),
- \( b = -48 \),
- \( c = -6 \).
5.
Substitute \( a \), \( b \), and \( c \) into the quadratic formula:
\[
t = \frac{-(-48) \pm \sqrt{(-48)^2 - 4(16)(-6)}}{2(16)}
\]
6.
Simplify step by step:
- Calculate \( -(-48) \):
\[
-(-48) = 48
\]
- Calculate \( (-48)^2 \):
\[
(-48)^2 = 2304
\]
- Calculate \( 4(16)(-6) \):
\[
4(16)(-6) = -384
\]
- Substitute these values into the discriminant \( b^2 - 4ac \):
\[
b^2 - 4ac = 2304 - (-384) = 2304 + 384 = 2688
\]
- Take the square root of the discriminant:
\[
\sqrt{2688} = \sqrt{16 \times 168} = 4\sqrt{168}
\]
Simplify \( \sqrt{168} \):
\[
\sqrt{168} = \sqrt{4 \times 42} = 2\sqrt{42}
\]
Therefore:
\[
\sqrt{2688} = 4 \times 2\sqrt{42} = 8\sqrt{42}
\]
- Substitute back into the quadratic formula:
\[
t = \frac{48 \pm 8\sqrt{42}}{32}
\]
7.
Simplify the expression:
Factor out the common factor of 8 in the numerator:
\[
t = \frac{8(6 \pm \sqrt{42})}{32}
\]
Simplify the fraction:
\[
t = \frac{6 \pm \sqrt{42}}{4}
\]
8.
Find the two solutions for \( t \):
- For the positive root:
\[
t_1 = \frac{6 + \sqrt{42}}{4}
\]
- For the negative root:
\[
t_2 = \frac{6 - \sqrt{42}}{4}
\]
Since time cannot be negative, we discard \( t_2 \).
9.
Calculate the approximate value of \( t_1 \):
First, approximate \( \sqrt{42} \):
\[
\sqrt{42} \approx 6.48
\]
Substitute this value into \( t_1 \):
\[
t_1 = \frac{6 + 6.48}{4} = \frac{12.48}{4} = 3.12
\]
10.
Round to the nearest tenth:
\[
t_1 \approx 3.1
\]
Final Answer:
The ball is in the air for approximately \(\boxed{3.1}\) seconds.
Parent Tip: Review the logic above to help your child master the concept of quadratic formula practice problems worksheet.
