### Instruction: The equation $y = -4.9t^2 + 23.8t$ describes the height (in meters) of a projectile launched from the ground at 23.8 meters per second. In how many seconds will the projectile first reach 28 meters in height?### Response: To find the time when the height is 28 meters, I need to substitute 28 for y and solve for t.
That gives me the equation $28 = -4.9t^2 + 23.8t$.
I can simplify this equation by subtracting 28 from both sides and dividing everything by -4.9.
That gives me $t^2 - 4.86t + 5.71 = 0$.
This is a quadratic equation, so I can use the quadratic formula to find the values of t that make it true.
The quadratic formula says that $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -4.86$, and $c = 5.71$.
Plugging in these values, I get $t = \frac{4.86 \pm \sqrt{(-4.86)^2 - 4(1)(5.71)}}{2(1)}$.
Simplifying, I get $t = \frac{4.86 \pm \sqrt{23.6196 - 22.84}}{2}$.
Simplifying further, I get $t = \frac{4.86 \pm \sqrt{0.7796}}{2}$.
Taking the square root of 0.7796, I get about 0.883.
So the possible values of t are $t = \frac{4.86 + 0.883}{2}$ or $t = \frac{4.86 - 0.883}{2}$.
Simplifying, I get $t \approx 2.87$ or $t \approx 1.99$.
These are the times when the height is 28 meters, but the problem asks for the first time, so I need to pick the smaller value of t.
Therefore, the answer is $t \approx 1.99$ seconds.