Consider the following descriptions of the vertical motion of an object subject only to the acceleration due to gravity. (Note: the acceleration due to gravity is \( 9.8 \mathrm{~m} / \mathrm{s}^{2} \).) a. Find the velocity of the object for all relevant times. b. Find the position of the object for all relevant times. c. Find the time when the object reaches its highest point (What is the height?) d. Find the time when the object strikes the ground. A payload is released at an elevation of 500 m from a hot-air balloon that is rising at a rate of \( 10 \mathrm{~m} / \mathrm{s} \). a. \( \mathrm{v}(\mathrm{t})= \)

To find the velocity of the object for all relevant times, we start by considering that the initial velocity \( v_0 \) is \( 10 \, \text{m/s} \) (upward) and the acceleration due to gravity \( g \) is \( 9.8 \, \text{m/s}^2 \) (downward). Therefore, the equation for velocity as a function of time \( v(t) \) is given by: \[ v(t) = v_0 - g t = 10 - 9.8t \] This means that after \( t \) seconds, the velocity of the payload will decrease due to gravity at a rate of \( 9.8 \, \text{m/s}^2 \). For the position of the object, we can use the kinematic equation, which incorporates the initial position \( y_0 = 500\, \text{m} \) and takes into account the initial velocity and gravity: \[ y(t) = y_0 + v_0 t - \frac{1}{2} g t^2 = 500 + 10t - 4.9t^2 \] This quadratic equation represents the height of the payload at any time \( t \). As time progresses, the object will move upward initially before it begins its descent. To find the time when the object reaches its highest point, we can set the velocity to zero: \[ 0 = 10 - 9.8 t \rightarrow t = \frac{10}{9.8} \approx 1.02\, \text{s} \] Substituting \( t \) back into the position equation gives us the maximum height: \[ y(1.02) = 500 + 10(1.02) - 4.9(1.02)^2 \approx 505.1\, \text{m} \] To determine when the object strikes the ground, we will set \( y(t) = 0 \) and solve for \( t \): \[ 0 = 500 + 10t - 4.9t^2 \] Rearranging gives a standard quadratic form: \[ 4.9t^2 - 10t - 500 = 0 \] Applying the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 4.9 \cdot (-500)}}{2 \cdot 4.9} \] Calculating gives a positive time when the object will strike the ground. Wouldn’t it be cool to visualize this in a fun way? Imagine a scenario where this payload is a bouncing ball! When dropped (or launched), you can play with a timer and see how high it goes or how quickly it comes back down! Physics really comes alive when you think about it in everyday fun contexts. And speaking of fun contexts, did you know that hunters in ancient civilizations used similar principles when they calculated the trajectories of their arrows and thrown spears? They would have instinctual knowledge about gravity's effect and adjust their aim accordingly. Think of the evolution of that knowledge into today’s breathtaking Olympic sports! It's all a big, fascinating cycle of motion!