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 \( 26 \mathrm{~m} / \mathrm{s} \).

To solve the problem, we will follow the steps outlined in the prompt. ### Given Information: - Acceleration due to gravity: \( g = -9.8 \, \text{m/s}^2 \) - Initial velocity of the object: \( v(0) = 26 \, \text{m/s} \) ### a. Find the velocity of the object for all relevant times. The acceleration \( a(t) \) is the derivative of the velocity \( v(t) \): \[ a(t) = v'(t) = g \] Integrating the acceleration to find the velocity: \[ v(t) = \int g \, dt = gt + C \] Where \( C \) is the constant of integration. Since the initial velocity \( v(0) = 26 \, \text{m/s} \), we can find \( C \): \[ v(0) = g(0) + C \implies 26 = 0 + C \implies C = 26 \] Thus, the velocity function is: \[ v(t) = -9.8t + 26 \] ### b. Find the position of the object for all relevant times. The position \( s(t) \) is the integral of the velocity: \[ s(t) = \int v(t) \, dt = \int (-9.8t + 26) \, dt \] Calculating the integral: \[ s(t) = -4.9t^2 + 26t + D \] Where \( D \) is the constant of integration. Since the object starts from the ground, \( s(0) = 0 \): \[ s(0) = -4.9(0)^2 + 26(0) + D \implies 0 = 0 + D \implies D = 0 \] Thus, the position function is: \[ s(t) = -4.9t^2 + 26t \] ### c. Find the time when the object reaches its highest point. What is the height? The object reaches its highest point when the velocity is zero: \[ v(t) = 0 \implies -9.8t + 26 = 0 \] Solving for \( t \): \[ 9.8t = 26 \implies t = \frac{26}{9.8} \] Calculating \( t \): \[ t \approx 2.6531 \, \text{s} \] Now, we find the height at this time: \[ s\left(\frac{26}{9.8}\right) = -4.9\left(\frac{26}{9.8}\right)^2 + 26\left(\frac{26}{9.8}\right) \] Calculating \( s(t) \): \[ s\left(\frac{26}{9.8}\right) \approx -4.9 \cdot 2.6531^2 + 26 \cdot 2.6531 \] ### d. Find the time when the object strikes the ground. The object strikes the ground when \( s(t) = 0 \): \[ -4.9t^2 + 26t = 0 \] Factoring out \( t \): \[ t(-4.9t + 26) = 0 \] This gives us \( t = 0 \) (initial time) or: \[ -4.9t + 26 = 0 \implies 4.9t = 26 \implies t = \frac{26}{4.9} \] Calculating \( t \): \[ t \approx 5.3061 \, \text{s} \] Now, let's calculate the height at the highest point and the time when the object strikes the ground. ### Calculating the height at the highest point: \[ s\left(\frac{26}{9.8}\right) = -4.9\left(\frac{26}{9.8}\right)^2 + 26\left(\frac{26}{9.8}\right) \] Calculating \( \frac{26}{9.8} \) and substituting: \[ \frac{26}{9.8} \approx 2.6531 \] Now substituting back into the position function: \[ s(2.6531) = -4.9(2.6531^2) + 26(2.6531) \] Let's perform the calculations for the height and the time when the object strikes the ground. ### Performing the calculations: 1. Calculate \( s(2.6531) \) 2. Calculate \( t \) when the object strikes the ground. Let's do these calculations. Calculate the value by following steps: - step0: Calculate: \(\frac{26}{4.9}\) - step1: Convert the expressions: \(\frac{26}{\frac{49}{10}}\) - step2: Multiply by the reciprocal: \(26\times \frac{10}{49}\) - step3: Multiply: \(\frac{26\times 10}{49}\) - step4: Multiply: \(\frac{260}{49}\) Calculate or simplify the expression \( -4.9*(26/9.8)^2 + 26*(26/9.8) \). Calculate the value by following steps: - step0: Calculate: \(-4.9\left(\frac{26}{9.8}\right)^{2}+26\left(\frac{26}{9.8}\right)\) - step1: Divide the numbers: \(-4.9\left(\frac{130}{49}\right)^{2}+26\left(\frac{26}{9.8}\right)\) - step2: Divide the numbers: \(-4.9\left(\frac{130}{49}\right)^{2}+26\times \frac{130}{49}\) - step3: Multiply the numbers: \(-\frac{1690}{49}+26\times \frac{130}{49}\) - step4: Multiply the numbers: \(-\frac{1690}{49}+\frac{3380}{49}\) - step5: Transform the expression: \(\frac{-1690+3380}{49}\) - step6: Add the numbers: \(\frac{1690}{49}\) ### Results of the Calculations: 1. **Height at the highest point:** \[ s\left(\frac{26}{9.8}\right) \approx 34.49 \, \text{m} \] 2. **Time when the object strikes the ground:** \[ t \approx 5.31 \, \text{s} \] ### Summary of Results: - **Velocity function:** \[ v(t) = -9.8t + 26 \] - **Position function:** \[ s(t) = -4.9t^2 + 26t \] - **Time to reach the highest point:** \[ t \approx 2.65 \, \text{s} \] - **Height at the highest point:** \[ \text{Height} \approx 34.49 \, \text{m} \] - **Time when the object strikes the ground:** \[ t \approx 5.31 \, \text{s} \] These results provide a complete analysis of the vertical motion of the object under the influence of gravity.