A ball is thrown upward from the top of a 200 foot tall building with a velocity of 40 feet per second. Take the positive direction upward and the origin of the coordinate system at ground level. What is the initial value problem for the position, \( x(t) \), of the ball at time \( t \) ? Select the correct answer. \( \frac{d^{2} x}{d t^{2}}=40, x(0)=200, \frac{d x}{d t}(0)=40 \) \( \frac{d^{2} x}{d t^{2}}=-40, x(0)=200, \frac{d x}{d t}(0)=40 \) \( \frac{d^{2} x}{d t^{2}}=32, x(0)=200, \frac{d x}{d t}(0)=40 \) \( \frac{d^{2} x}{d t^{2}}=-32, x(0)=200, \frac{x A}{d t}(0)=40 \) \( \frac{d^{2} x}{d t^{2}}=200, x(0)=32, \frac{d x}{d t}(0)=40 \)

To formulate the initial value problem for the position \( x(t) \) of the ball, we need to consider the following: 1. **Acceleration due to gravity**: The ball is thrown upward, and the only force acting on it after it is thrown is gravity, which acts downward. The acceleration due to gravity is approximately \( 32 \, \text{ft/s}^2 \) (downward). Therefore, the second derivative of the position function, which represents acceleration, will be negative: \[ \frac{d^2 x}{dt^2} = -32 \] 2. **Initial position**: The ball is thrown from the top of a 200-foot tall building, so the initial position at time \( t = 0 \) is: \[ x(0) = 200 \] 3. **Initial velocity**: The ball is thrown upward with an initial velocity of 40 feet per second, so the initial velocity at time \( t = 0 \) is: \[ \frac{dx}{dt}(0) = 40 \] Now, we can summarize the initial value problem for the position \( x(t) \): \[ \frac{d^2 x}{dt^2} = -32, \quad x(0) = 200, \quad \frac{dx}{dt}(0) = 40 \] Thus, the correct answer is: \[ \frac{d^{2} x}{d t^{2}}=-32, \, x(0)=200, \, \frac{d x}{d t}(0)=40 \]