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. \( \frac{d^{2} x}{d t^{2}}=40, x(0)=200, \frac{d x}{d t}(0)=40 \) \( \frac{d^{2} x}{d t^{2}}=-40, x(0)=200, \frac{d x}{d t}(0)=40 \) \( \frac{d^{2} x}{d t^{2}}=32, x(0)=200, \frac{d x}{d t}(0)=40 \) \( \frac{d^{2} x}{d t^{2}}=-32, x(0)=200, \frac{x A}{d t}(0)=40 \) \( \frac{d^{2} x}{d t^{2}}=200, x(0)=32, \frac{d x}{d t}(0)=40 \)
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When throwing a ball upward, gravity acts as a constant force in the negative direction, leading to a downward acceleration of approximately \(32 \, \text{ft/s}^2\). Thus, the second derivative of position with respect to time should be negative, which aligns with the motion of the ball. Additionally, the initial height of the ball is \(200\) feet, and the initial velocity is \(40\) feet per second. So, the correct initial value problem for the position of the ball at time \(t\) is: \(\frac{d^{2} x}{d t^{2}}=-32, x(0)=200, \frac{d x}{d t}(0)=40\). It's important to remember that when analyzing motion in physics, always consider how forces like gravity influence the path of a moving object. This foundational concept links to a variety of practical applications, from sports to engineering designs. Additionally, a common mistake people make in problems like this is not accounting for the downward acceleration correctly. Ensure you keep track of the signs associated with the forces acting on the object! This will help you avoid confusion in more complex physics applications too!
