The problem provides two key functions:

1. The height of the projectile at time $t$, given by: $h(t) = -16t^2 + v_0 t + y_0$ where:

   • $v_0 = 100 \, \text{ft/s}$ (initial vertical velocity),
   • $y_0 = 5 \, \text{ft}$ (initial height from the ground).

2. The horizontal distance of the projectile at time $t$, given by: $d(t) = v_1 t$ where:

   • $v_1 = 115 \, \text{ft/s}$ (initial horizontal velocity).

To find:

1. Height of the projectile after 3 seconds: $h(3) = -16(3)^2 + 100(3) + 5$

2. Horizontal distance from the launcher after 3 seconds: $d(3) = 115(3)$

Let me calculate both values.After 3 seconds:

1. The height of the projectile is $161 \, \text{feet}$.
2. The horizontal distance from the launcher is $345 \, \text{feet}$.

Would you like further details or explanations on any part of this?

Here are 5 related questions you might explore:

1. What would happen to the height if the initial vertical velocity was different?
2. How long would the projectile remain in the air before hitting the ground?
3. How far horizontally would the projectile travel before landing?
4. How does the acceleration due to gravity (-16 ft/s²) affect the height over time?
5. What if the projectile was launched from a higher platform, how would that impact the motion?

Tip: When dealing with projectile motion, splitting the problem into vertical and horizontal components helps simplify the analysis.