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 \( 13 \mathrm{~m} / \mathrm{s} \).

Certainly! Let's analyze the vertical motion of the payload released from the hot-air balloon step by step. We'll consider each part of the problem (a to d) systematically. --- ## **Given:** - **Initial Elevation (\( y_0 \))**: 400 meters - **Initial Velocity (\( v_0 \))**: 13 meters/second (upward) - **Acceleration due to Gravity (\( g \))**: \( 9.8 \, \text{m/s}^2 \) (acting downward) *For clarity, we'll define the upward direction as positive.* --- ## **a. Find the Velocity of the Object for All Relevant Times** The velocity of an object under constant acceleration is given by the equation: \[ v(t) = v_0 + a t \] Where: - \( v(t) \) is the velocity at time \( t \). - \( v_0 \) is the initial velocity. - \( a \) is the acceleration (in this case, \( a = -g = -9.8 \, \text{m/s}^2 \) since gravity acts downward). **Calculation:** \[ v(t) = 13 \, \text{m/s} - 9.8 \, \text{m/s}^2 \times t \] \[ \boxed{v(t) = 13 - 9.8\, t \quad \text{meters/second}} \] --- ## **b. Find the Position of the Object for All Relevant Times** The position of an object under constant acceleration is given by the equation: \[ y(t) = y_0 + v_0 t + \frac{1}{2} a t^2 \] Where: - \( y(t) \) is the position at time \( t \). - \( y_0 \) is the initial position. **Calculation:** \[ y(t) = 400 \, \text{m} + 13 \, \text{m/s} \times t - \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times t^2 \] \[ y(t) = 400 + 13t - 4.9 t^2 \] \[ \boxed{y(t) = 400 + 13\, t - 4.9\, t^2 \quad \text{meters}} \] --- ## **c. Find the Time When the Object Reaches Its Highest Point and Its Maximum Height** ### **Time to Reach Maximum Height (\( t_{\text{max}} \)):** At the highest point, the vertical velocity becomes zero (\( v(t_{\text{max}}) = 0 \)). Using the velocity equation: \[ 0 = 13 - 9.8\, t_{\text{max}} \] \[ 9.8\, t_{\text{max}} = 13 \] \[ t_{\text{max}} = \frac{13}{9.8} \approx 1.3265 \, \text{seconds} \] \[ \boxed{t_{\text{max}} \approx 1.33 \, \text{seconds}} \] ### **Maximum Height (\( y_{\text{max}} \)):** Plug \( t_{\text{max}} \) into the position equation: \[ y_{\text{max}} = 400 + 13 \times 1.3265 - 4.9 \times (1.3265)^2 \] \[ y_{\text{max}} = 400 + 17.1445 - 4.9 \times 1.758 \] \[ y_{\text{max}} = 400 + 17.1445 - 8.63 \approx 408.51 \, \text{meters} \] \[ \boxed{y_{\text{max}} \approx 408.5 \, \text{meters}} \] *Alternatively, using the formula \( y_{\text{max}} = y_0 + \frac{v_0^2}{2g} \):* \[ y_{\text{max}} = 400 + \frac{13^2}{2 \times 9.8} = 400 + \frac{169}{19.6} \approx 400 + 8.63 = 408.63 \, \text{meters} \] --- ## **d. Find the Time When the Object Strikes the Ground** The object strikes the ground when its position \( y(t) = 0 \). We'll solve the position equation for \( t \): \[ 0 = 400 + 13t - 4.9 t^2 \] Rearranging: \[ 4.9 t^2 - 13t - 400 = 0 \] This is a quadratic equation of the form \( at^2 + bt + c = 0 \), where: - \( a = 4.9 \) - \( b = -13 \) - \( c = -400 \) **Using the Quadratic Formula:** \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] **Plugging in the values:** \[ t = \frac{13 \pm \sqrt{(-13)^2 - 4 \times 4.9 \times (-400)}}{2 \times 4.9} \] \[ t = \frac{13 \pm \sqrt{169 + 7840}}{9.8} \] \[ t = \frac{13 \pm \sqrt{8009}}{9.8} \] \[ \sqrt{8009} \approx 89.5 \] **Calculating the two possible solutions:** 1. **Positive Root:** \[ t = \frac{13 + 89.5}{9.8} = \frac{102.5}{9.8} \approx 10.45 \, \text{seconds} \] 2. **Negative Root:** \[ t = \frac{13 - 89.5}{9.8} = \frac{-76.5}{9.8} \approx -7.81 \, \text{seconds} \quad \text{(Discarded, as time cannot be negative)} \] \[ \boxed{t \approx 10.45 \, \text{seconds}} \] **Conclusion:** The payload will strike the ground approximately **10.45 seconds** after being released. --- ## **Summary of Results:** - **a. Velocity Function:** \( v(t) = 13 - 9.8\, t \) meters/second - **b. Position Function:** \( y(t) = 400 + 13\, t - 4.9\, t^2 \) meters - **c. Time to Highest Point:** \( \approx 1.33 \) seconds **Maximum Height:** \( \approx 408.5 \) meters - **d. Time to Strike Ground:** \( \approx 10.45 \) seconds Feel free to ask if you need further clarification on any of these steps!