Let's break down the problem step by step.

Problem Breakdown:

1. Position Function: The position of the projectile is given by: $s(t) = -16t^2 + 224t + 240$ for $0 \leq t \leq 15$.

2. Graphing the Position Function: You need to graph the function $s(t)$ over the interval $0 \leq t \leq 15$. The correct graph will likely be a parabola opening downwards (since the coefficient of $t^2$ is negative), starting from a positive position value.

3. Instantaneous Velocity Equals Zero: The instantaneous velocity is zero at the vertex of the parabola, which occurs at the maximum height of the projectile. You can find this by determining the time $t$ at which the derivative $s'(t)$ is zero.

4. Secant Line Slopes: Finally, the problem asks you to approximate the instantaneous velocity at $t = a$ (which is likely the time when velocity equals zero) by calculating the slopes of secant lines over increasingly small intervals around this time.

Steps to Solve:

Part (a): Graphing the Position Function

The function $s(t) = -16t^2 + 224t + 240$ is a quadratic function, and its graph is a parabola. Since the coefficient of $t^2$ is negative, the parabola opens downwards.

The correct graph should show a parabola that opens downwards, with the vertex at the time $t$ when the projectile reaches its maximum height.

Part (b): Finding the Time When Velocity is Zero

1. Velocity Function: The velocity is the derivative of the position function: $v(t) = s'(t) = -32t + 224$
2. Set Velocity to Zero: Set $v(t) = 0$ and solve for $t$: 32t = 224 \\ t = 7$$ So, the time when the velocity is zero is $$t = 7$$.

Part (c): Approximating Instantaneous Velocity Using Secant Lines

For each interval, calculate the slope of the secant line as: $\text{Slope of secant line} = \frac{s(a + h) - s(a)}{h}$ where $a = 7$ and $h$ is the interval size.

1. Interval [7, 7.5]: $h = 0.5, \quad \text{Slope} = \frac{s(7.5) - s(7)}{0.5}$
2. Interval [7, 7.1]: $h = 0.1, \quad \text{Slope} = \frac{s(7.1) - s(7)}{0.1}$
3. Interval [7, 7.01]: $h = 0.01, \quad \text{Slope} = \frac{s(7.01) - s(7)}{0.01}$
4. Interval [7, 7.001]: $h = 0.001, \quad \text{Slope} = \frac{s(7.001) - s(7)}{0.001}$
5. Interval [7, 7.0001]: $h = 0.0001, \quad \text{Slope} = \frac{s(7.0001) - s(7)}{0.0001}$

You can calculate these slopes using the position function values.

Would you like me to calculate the specific slope values for these intervals?