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.)

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### Given:
- Acceleration due to gravity: $$ g = -9.8 \, \text{m/s}^2 $$
- Initial velocity: $$ v(0) = 25 \, \text{m/s} $$

### 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 $$ v(0) = 25 \, \text{m/s} $$, we can find $$ C $$:
$$
v(0) = g(0) + C \implies 25 = 0 + C \implies C = 25
$$
Thus, the velocity function is:
$$
v(t) = -9.8t + 25
$$

### b. Find the position of the object for all relevant times.

The position $$ s(t) $$ is the integral of the velocity:
$$
s(t) = \int v(t) \, dt = \int (-9.8t + 25) \, dt
$$
Calculating the integral:
$$
s(t) = -4.9t^2 + 25t + C
$$
Where $$ C $$ is the constant of integration. Since the initial position $$ s(0) = 0 $$:
$$
s(0) = -4.9(0)^2 + 25(0) + C \implies 0 = 0 + C \implies C = 0
$$
Thus, the position function is:
$$
s(t) = -4.9t^2 + 25t
$$

### 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:
$$
v(t) = 0 \implies -9.8t + 25 = 0
$$
Solving for $$ t $$:
$$
9.8t = 25 \implies t = \frac{25}{9.8}
$$
Calculating $$ t $$:
$$
t \approx 2.55 \, \text{s}
$$

Now, we find the height at this time:
$$
s(2.55) = -4.9(2.55)^2 + 25(2.55)
$$
Calculating $$ s(2.55) $$:
$$
s(2.55) \approx -4.9(6.5025) + 63.75 \approx -31.87 + 63.75 \approx 31.88 \, \text{m}
$$

### d. Find the time when the object strikes the ground.

The object strikes the ground when $$ s(t) = 0 $$:
$$
-4.9t^2 + 25t = 0
$$
Factoring out $$ t $$:
$$
t(-4.9t + 25) = 0
$$
This gives us two solutions:
1. $$ t = 0 $$ (initial position)
2. $$ -4.9t + 25 = 0 \implies t = \frac{25}{4.9} $$

Calculating $$ t $$:
$$
t \approx 5.10 \, \text{s}
$$

### Summary of Results:
a. $$ v(t) = -9.8t + 25 $$

b. $$ s(t) = -4.9t^2 + 25t $$

c. The object's highest point is $$ 31.88 \, \text{m} $$ at time $$ t \approx 2.55 \, \text{s} $$.

d. $$ t \approx 5.10 \, \text{s} $$.
a. $$ v(t) = -9.8t + 25 $$ b. $$ s(t) = -4.9t^2 + 25t $$ c. The object reaches its highest point at $$ t \approx 2.55 \, \text{s} $$ with a height of $$ 31.88 \, \text{m} $$. d. The object strikes the ground at $$ t \approx 5.10 \, \text{s} $$.

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