Consider the following descriptions of the vertical motion of an object subject only to the acceleration due to gravity. Begin with the acceleration equation \( \mathrm{a}(\mathrm{t})=\mathrm{v}^{\prime}(\mathrm{t})=\mathrm{g} \), where \( \mathrm{g}=-9.8 \mathrm{~m} / \mathrm{s}^{2} \). 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 softball is popped up vertically (from the ground) with a velocity of \( 25 \mathrm{~m} / \mathrm{s} \). a. \( v(\mathrm{t})=\square \) b. \( \mathrm{s}(\mathrm{t})=\square \) c. The object's highest point is \( \square \mathrm{m} \) at time \( \mathrm{t}=\square \mathrm{s} \). (Simplify your answers. Round to two decimal places as needed.) d. \( \mathrm{t}=\square \) (Simplify your answer. Round to two decimal places as needed.)

To solve this problem, we can break it down step by step using the equations of motion under constant acceleration. ### a. Find the velocity of the object for all relevant times. The velocity \( v(t) \) can be calculated by integrating the acceleration function: \[ v(t) = v_0 + \int g \, dt = v_0 + gt \] Given that the initial velocity \( v_0 = 25 \, \text{m/s} \) and \( g = -9.8 \, \text{m/s}^2 \): \[ v(t) = 25 - 9.8t \] ### b. Find the position of the object for all relevant times. To find the position \( s(t) \), we can integrate the velocity function: \[ s(t) = s_0 + \int v(t) \, dt = s_0 + \int (25 - 9.8t) \, dt \] Assuming the initial position \( s_0 = 0 \): \[ s(t) = 0 + 25t - 4.9t^2 \] Thus, \[ s(t) = 25t - 4.9t^2 \] ### c. Find the time when the object reaches its highest point. What is the height? The highest point occurs when the velocity becomes zero: \[ 0 = 25 - 9.8t \rightarrow t = \frac{25}{9.8} \approx 2.55 \text{ s} \] To find the height at this time: \[ s(2.55) = 25(2.55) - 4.9(2.55^2) \approx 31.88 \text{ m} \] ### d. Find the time when the object strikes the ground. The object strikes the ground when \( s(t) = 0 \): \[ 0 = 25t - 4.9t^2 \rightarrow 4.9t^2 - 25t = 0 \] Factoring: \[ t(4.9t - 25) = 0 \rightarrow t = 0 \text{ or } t = \frac{25}{4.9} \approx 5.10 \text{ s} \] Now let's summarize our results: - a. \( v(t) = 25 - 9.8t \) - b. \( s(t) = 25t - 4.9t^2 \) - c. The object's highest point is \( 31.88 \, \text{m} \) at time \( t = 2.55 \, \text{s} \). - d. \( t \approx 5.10 \, \text{s} \) when it strikes the ground.