The given problem provides the equation for a projectile's height as:

$h(t) = -16t^2 + v_0 t + h_0$

Where:

• $t$ is the time in seconds,
• $v_0$ is the initial velocity,
• $h_0$ is the initial height,
• The coefficient $-16$ represents the effect of gravity in feet per second squared.

The question asks to determine when a tennis ball, served from a height of 2 feet with an initial speed of 130 feet per second, reaches its maximum height.

Step-by-Step Solution:

Given values:

• $h_0 = 2 \, \text{feet}$ (initial height),
• $v_0 = 130 \, \text{feet/second}$ (initial velocity).

The equation simplifies to:

$h(t) = -16t^2 + 130t + 2$

The time at which the ball reaches its maximum height can be found using the formula for the vertex of a parabola. For a quadratic equation $h(t) = at^2 + bt + c$, the time $t$ at which the maximum height occurs is given by:

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

In this case:

• $a = -16$,
• $b = 130$.

Substituting these into the formula:

$t = \frac{-130}{2(-16)} = \frac{-130}{-32} = 4.06 \, \text{seconds}$

Thus, the ball reaches its maximum height after approximately 4.06 seconds.

Would you like more details or have any questions?

Here are 5 related questions:

1. How do we calculate the maximum height attained by the ball?
2. What effect does changing the initial velocity have on the time to reach maximum height?
3. Can this equation be used for projectiles with different initial heights or velocities?
4. How would the equation change if gravity were different, such as on another planet?
5. What is the significance of the negative coefficient in the equation?

Tip: In projectile motion problems, the time to reach maximum height only depends on the initial vertical velocity and gravity, not the initial height.