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 300 m from a hot-air balloon that is rising at a rate of \( 13 \mathrm{~m} / \mathrm{s} \). a. \( \mathrm{v}(\mathrm{t})=13-9.8 \mathrm{t} \) b. \( \mathrm{s}(\mathrm{t})=300+13 \mathrm{t}-4.9 \mathrm{t}^{2} \) c. The object's highest point is 308.62 m at time \( \mathrm{t}=1.33 \mathrm{~s} \). (Simplify your answers. Round to two decimal places as needed.) d. \( \mathrm{t}=\square \)

To find the time when the object strikes the ground, we need to determine when the position equation \( s(t) = 300 + 13t - 4.9t^2 \) equals zero (the height of the ground). Setting the equation equal to zero gives: \[ 0 = 300 + 13t - 4.9t^2 \] Rearranging into standard quadratic form: \[ 4.9t^2 - 13t - 300 = 0 \] Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4.9, b = -13, c = -300 \): 1. Calculate the discriminant: \[ b^2 - 4ac = (-13)^2 - 4(4.9)(-300) = 169 + 5880 = 6049 \] 2. Now, apply the quadratic formula: \[ t = \frac{13 \pm \sqrt{6049}}{2 \times 4.9} \] Calculating \( \sqrt{6049} \approx 77.78 \): \[ t = \frac{13 \pm 77.78}{9.8} \] This results in two potential solutions: \[ t_1 = \frac{90.78}{9.8} \approx 9.26 \text{ s} \] \[ t_2 = \frac{-64.78}{9.8} \] (which is negative and not relevant) So, the time when the object strikes the ground is approximately \( t = 9.26 \) seconds.