Question
A ball Is thrown from an Inital height of 2 meters with an Inital upward velocity of 5 m/s. The ball's height h （In meters） after t seconds Is given by the
H=2+5t-5t^2
Find all values of t for which the ball's height is 3 meters.
H=2+5t-5t^2
Find all values of t for which the ball's height is 3 meters.
Answers
Answered by
GPT-4o mini
To find the values of \( t \) for which the ball's height \( h \) is 3 meters, we need to set the height equation equal to 3 and solve for \( t \):
\[
h = 2 + 5t - 5t^2 = 3
\]
Subtracting 3 from both sides gives us:
\[
2 + 5t - 5t^2 - 3 = 0
\]
Simplifying the equation results in:
\[
-5t^2 + 5t - 1 = 0
\]
To make calculations easier, we can multiply through by -1:
\[
5t^2 - 5t + 1 = 0
\]
Next, we will use the quadratic formula to solve for \( t \). The quadratic formula is given by:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 5 \), \( b = -5 \), and \( c = 1 \). We can plug in these values:
First, we calculate the discriminant \( b^2 - 4ac \):
\[
b^2 - 4ac = (-5)^2 - 4(5)(1) = 25 - 20 = 5
\]
Now we can substitute into the quadratic formula:
\[
t = \frac{-(-5) \pm \sqrt{5}}{2 \cdot 5} = \frac{5 \pm \sqrt{5}}{10}
\]
This gives us two possible values for \( t \):
\[
t_1 = \frac{5 + \sqrt{5}}{10} \quad \text{and} \quad t_2 = \frac{5 - \sqrt{5}}{10}
\]
Thus, the times \( t \) at which the ball's height is 3 meters are:
\[
t = \frac{5 + \sqrt{5}}{10} \quad \text{and} \quad t = \frac{5 - \sqrt{5}}{10}
\]
Calculating these values numerically, we find:
1. \( t_1 \approx \frac{5 + 2.236}{10} \approx \frac{7.236}{10} \approx 0.7236 \)
2. \( t_2 \approx \frac{5 - 2.236}{10} \approx \frac{2.764}{10} \approx 0.2764 \)
Therefore, the values of \( t \) for which the ball's height is 3 meters are approximately \( t \approx 0.724 \) seconds and \( t \approx 0.276 \) seconds.
\[
h = 2 + 5t - 5t^2 = 3
\]
Subtracting 3 from both sides gives us:
\[
2 + 5t - 5t^2 - 3 = 0
\]
Simplifying the equation results in:
\[
-5t^2 + 5t - 1 = 0
\]
To make calculations easier, we can multiply through by -1:
\[
5t^2 - 5t + 1 = 0
\]
Next, we will use the quadratic formula to solve for \( t \). The quadratic formula is given by:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 5 \), \( b = -5 \), and \( c = 1 \). We can plug in these values:
First, we calculate the discriminant \( b^2 - 4ac \):
\[
b^2 - 4ac = (-5)^2 - 4(5)(1) = 25 - 20 = 5
\]
Now we can substitute into the quadratic formula:
\[
t = \frac{-(-5) \pm \sqrt{5}}{2 \cdot 5} = \frac{5 \pm \sqrt{5}}{10}
\]
This gives us two possible values for \( t \):
\[
t_1 = \frac{5 + \sqrt{5}}{10} \quad \text{and} \quad t_2 = \frac{5 - \sqrt{5}}{10}
\]
Thus, the times \( t \) at which the ball's height is 3 meters are:
\[
t = \frac{5 + \sqrt{5}}{10} \quad \text{and} \quad t = \frac{5 - \sqrt{5}}{10}
\]
Calculating these values numerically, we find:
1. \( t_1 \approx \frac{5 + 2.236}{10} \approx \frac{7.236}{10} \approx 0.7236 \)
2. \( t_2 \approx \frac{5 - 2.236}{10} \approx \frac{2.764}{10} \approx 0.2764 \)
Therefore, the values of \( t \) for which the ball's height is 3 meters are approximately \( t \approx 0.724 \) seconds and \( t \approx 0.276 \) seconds.