The problem in the image involves a ball thrown directly upward with a given initial height and velocity. The height of the ball at time $t$ seconds after it is thrown is modeled by the function:

$s(t) = -16t^2 + 28t + 3$

We are asked to determine:

1. The time at which the ball reaches its maximum height.
2. The maximum height.

1. Time to reach maximum height

This is a quadratic function of the form $s(t) = at^2 + bt + c$. The time at which the maximum height is reached can be found using the formula for the vertex of a parabola:

$t = \frac{-b}{2a}$

In the given function, $a = -16$, $b = 28$, and $c = 3$. Plugging these values into the formula:

$t = \frac{-28}{2(-16)} = \frac{-28}{-32} = 0.875 \text{ seconds}$

2. Maximum height

To find the maximum height, substitute $t = 0.875$ back into the height equation $s(t)$:

$s(0.875) = -16(0.875)^2 + 28(0.875) + 3$

Calculating each term:

$s(0.875) = -16(0.765625) + 28(0.875) + 3$ $s(0.875) = -12.25 + 24.5 + 3$ $s(0.875) = 15.25 \text{ feet}$

Final Answer:

• The ball reaches its maximum height of 15.25 feet after 0.875 seconds.

Would you like more details or have any questions?

Here are 5 related questions for further understanding:

1. How do we derive the formula for the vertex of a quadratic function?
2. What are the physical meanings of the constants in the height function?
3. How would the equation change if the initial velocity were doubled?
4. What is the significance of the negative coefficient for $t^2$ in this context?
5. How can we find the time the ball hits the ground after reaching its peak?

Tip: Always check the units (feet, seconds) in these problems to ensure they make physical sense in context.