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 400 m from a hot-air balloon that is rising at a rate of \( 13 \mathrm{~m} / \mathrm{s} \).

Let's analyze the vertical motion of the payload step by step. We'll define upward as the positive direction. ### Given: - **Initial elevation (\( y_0 \))**: 400 meters - **Initial velocity (\( v_0 \))**: 13 m/s (upward, since the balloon is rising) - **Acceleration due to gravity (\( g \))**: \( 9.8 \, \text{m/s}^2 \) (downward) ### a. **Velocity of the Object for All Relevant Times** The velocity of the object as a function of time can be found using the equation of motion: \[ v(t) = v_0 + a t \] Here: - \( v_0 = 13 \, \text{m/s} \) (initial upward velocity) - \( a = -g = -9.8 \, \text{m/s}^2 \) (acceleration due to gravity, acting downward) So, \[ v(t) = 13 \, \text{m/s} - 9.8 \, \text{m/s}^2 \cdot t \] **Final Expression:** \[ v(t) = 13 - 9.8\, t \quad \text{m/s} \] ---