The height $h(t)$ of a projectile launched from the ground is given by the equation:

$h(t) = -16t^2 + v_0t$

where:

• $h(t)$ is the height in feet at time $t$ seconds,
• $v_0$ is the initial velocity in feet per second,
• $-16t^2$ is due to the acceleration of gravity (in feet per second squared).

You are tasked with solving two things:

• (a) When the projectile reaches a specific height $h$.
• (b) When the projectile returns to the ground (i.e., when $h = 0$).

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(a) Find the time(s) when the projectile reaches a height of $h$ feet

We will solve for $t$ when $h(t) = h$:

$h = -16t^2 + v_0 t$

Rearranging this quadratic equation:

$16t^2 - v_0 t + h = 0$

To find the time $t$, use the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 16$, $b = -v_0$, and $c = h$.

The two solutions represent the times at which the projectile reaches the given height.

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(b) Find the time when the projectile returns to the ground

To find when the projectile returns to the ground, set $h(t) = 0$:

$0 = -16t^2 + v_0 t$

This equation can be factored:

$0 = t(-16t + v_0)$

Thus, the solutions are:

$t = 0 \quad \text{(when the projectile is launched)}, \quad \text{or} \quad t = \frac{v_0}{16}$

The time $t = \frac{v_0}{16}$ is when the projectile returns to the ground.

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Would you like more detailed steps for solving part (a), or do you have specific values you'd like to calculate?

Here are 5 related questions:

1. How would changing the initial velocity affect the maximum height of the projectile?
2. How can you calculate the maximum height the projectile reaches?
3. What would happen if you included air resistance in this model?
4. How would you find the time at which the projectile reaches its maximum height?
5. How can you extend this analysis to projectiles launched at an angle?

Tip: The time when the projectile reaches its maximum height occurs at $t = \frac{v_0}{32}$, since gravity reduces the velocity to zero at that instant.