A projectile is launched from ground level with an initial velocity of  feet per second. Neglecting air  resistance, its height in feet t seconds after launch is given by . Find the  time(s) that the projectile will  (a) reach a height of  ft and  (b) return to the ground when  is  feet per second. v 0 s= −16t +v t 2 0 

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.

---

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.

A projectile is launched from ground level with an initial velocity of feet per second. Neglecting air resistance, its height in feet t seconds after launch is given by . Find the time(s) that the projectile will (a) reach a height of ft and (b) return to the ground when is feet per second.

Solve a projectile motion problem involving quadratic equations. Learn how to calculate the time when the projectile reaches a specific height and when it returns to the ground. This solution uses the quadratic formula and is suitable for students in Grades 10-12.

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