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 \( 10 \mathrm{~m} / \mathrm{s} \).

Let's analyze the vertical motion of the payload released from the hot-air balloon considering the acceleration due to gravity. We'll address each part of the problem step by step. --- ### **Given:** - **Initial Elevation (\( y_0 \))**: 500 meters - **Initial Velocity (\( v_0 \))**: Since the balloon is rising at \( 10 \, \text{m/s} \), the payload is released with an initial upward velocity of \( v_0 = +10 \, \text{m/s} \). - **Acceleration due to Gravity (\( a \))**: \( -9.8 \, \text{m/s}^2 \) (negative sign indicates downward acceleration) We'll define the upward direction as positive. --- ### **a. Velocity of the Object for All Relevant Times** **Formula:** \[ v(t) = v_0 + a t \] **Calculation:** \[ v(t) = 10 \, \text{m/s} + (-9.8 \, \text{m/s}^2) \cdot t \] \[ v(t) = 10 - 9.8t \, \text{m/s} \] **Result:** \[ \boxed{v(t) = 10\, \text{m/s} - 9.8\, t\, \text{m/s}} \] --- ### **b. Position of the Object for All Relevant Times** **Formula:** \[ y(t) = y_0 + v_0 t + \frac{1}{2} a t^2 \] **Calculation:** \[ y(t) = 500\, \text{m} + (10\, \text{m/s}) \cdot t + \frac{1}{2} \cdot (-9.8\, \text{m/s}^2) \cdot t^2 \] \[ y(t) = 500 + 10t - 4.9t^2 \, \text{m} \] **Result:** \[ \boxed{y(t) = 500\, \text{m} + 10\, t\, \text{m/s} - 4.9\, t^2\, \text{m}} \] --- ### **c. Time and Height When the Object Reaches Its Highest Point** At the highest point, the vertical velocity becomes zero. **Finding Time (\( t_{\text{max}} \)) When \( v(t) = 0 \):** \[ 0 = 10 - 9.8t \] \[ 9.8t = 10 \] \[ t_{\text{max}} = \frac{10}{9.8} \approx 1.0204 \, \text{seconds} \] **Finding Height (\( y_{\text{max}} \)) at \( t_{\text{max}} \):** \[ y_{\text{max}} = 500 + 10 \cdot 1.0204 - 4.9 \cdot (1.0204)^2 \] \[ y_{\text{max}} \approx 500 + 10.204 - 4.9 \cdot 1.0412 \] \[ y_{\text{max}} \approx 500 + 10.204 - 5.1024 \] \[ y_{\text{max}} \approx 505.1016 \, \text{meters} \] **Results:** - **Time to Reach Highest Point:** \[ \boxed{t_{\text{max}} \approx 1.02\, \text{seconds}} \] - **Height at Highest Point:** \[ \boxed{y_{\text{max}} \approx 505.10\, \text{meters}} \] --- ### **d. Time When the Object Strikes the Ground** When the object strikes the ground, its position \( y(t) = 0 \). **Setting Position Equation to Zero:** \[ 0 = 500 + 10t - 4.9t^2 \] \[ 4.9t^2 - 10t - 500 = 0 \] **Solving the Quadratic Equation \( 4.9t^2 - 10t - 500 = 0 \):** Using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where: - \( a = 4.9 \) - \( b = -10 \) - \( c = -500 \) **Calculations:** \[ t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 4.9 \cdot (-500)}}{2 \cdot 4.9} \] \[ t = \frac{10 \pm \sqrt{100 + 9800}}{9.8} \] \[ t = \frac{10 \pm \sqrt{9900}}{9.8} \] \[ \sqrt{9900} \approx 99.497 \] \[ t = \frac{10 \pm 99.497}{9.8} \] **Valid Solution (Positive Time):** \[ t = \frac{10 + 99.497}{9.8} \] \[ t \approx \frac{109.497}{9.8} \] \[ t \approx 11.184 \, \text{seconds} \] **Result:** \[ \boxed{t \approx 11.18\, \text{seconds}} \] --- ### **Summary of Results:** - **a. Velocity as a Function of Time:** \[ v(t) = 10\, \text{m/s} - 9.8\, t\, \text{m/s} \] - **b. Position as a Function of Time:** \[ y(t) = 500\, \text{m} + 10\, t\, \text{m/s} - 4.9\, t^2\, \text{m} \] - **c. Time and Height at Highest Point:** \[ t_{\text{max}} \approx 1.02\, \text{seconds} \] \[ y_{\text{max}} \approx 505.10\, \text{meters} \] - **d. Time When the Object Strikes the Ground:** \[ t \approx 11.18\, \text{seconds} \]