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 $$ 25 \mathrm{~m} / \mathrm{s} $$.

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Let's address **Part c** of the problem, which involves finding the time when the object reaches its highest point and determining that maximum height.

### **c. Finding the Time to Reach the Highest Point and the Maximum Height**

**Given:**
- Initial velocity, $$ v_0 = 25 \, \text{m/s} $$ (upwards)
- Acceleration due to gravity, $$ g = -9.8 \, \text{m/s}^2 $$ (downwards)

**1. Time to Reach the Highest Point ($$ t_{\text{max}} $$)**

At the highest point, the velocity of the object becomes zero. We can use the velocity equation derived from integrating the acceleration:

$$
v(t) = v_0 + a t
$$

Setting $$ v(t) = 0 $$ to find $$ t_{\text{max}} $$:

$$
0 = 25 \, \text{m/s} - 9.8 \, \text{m/s}^2 \cdot t_{\text{max}}
$$

$$
9.8 \, \text{m/s}^2 \cdot t_{\text{max}} = 25 \, \text{m/s}
$$

$$
t_{\text{max}} = \frac{25}{9.8} \approx 2.55 \, \text{seconds}
$$

**2. Maximum Height ($$ h_{\text{max}} $$)**

We can find the maximum height using the position equation derived from integrating the velocity:

$$
s(t) = s_0 + v_0 t + \frac{1}{2} a t^2
$$

Assuming the object is launched from the ground ($$ s_0 = 0 $$):

$$
h_{\text{max}} = 25 \cdot t_{\text{max}} + \frac{1}{2} (-9.8) t_{\text{max}}^2
$$

Plugging in $$ t_{\text{max}} $$:

$$
h_{\text{max}} = 25 \cdot 2.55 + \frac{1}{2} (-9.8) \cdot (2.55)^2
$$

$$
h_{\text{max}} = 63.75 - 31.89 \approx 31.86 \, \text{meters}
$$

**Alternatively**, using the energy-based formula for maximum height:

$$
h_{\text{max}} = \frac{v_0^2}{2 |g|} = \frac{25^2}{2 \cdot 9.8} = \frac{625}{19.6} \approx 31.89 \, \text{meters}
$$

### **Final Answer**

**c.** The object reaches its highest point after approximately 2.55 seconds, attaining a maximum height of about 31.9 meters.
The object reaches its highest point after about 2.55 seconds and attains a maximum height of approximately 31.9 meters.

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