To find the rock's displacement after 1.00 second, we can use the kinematic equation for uniformly accelerated motion:

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

Where:

• $s$ is the displacement (what we want to find),
• $v_0 = 24.5 \, \text{m/s}$ is the initial velocity,
• $a = -9.81 \, \text{m/s}^2$ is the acceleration due to gravity (negative because it acts downward),
• $t = 1.00 \, \text{s}$ is the time.

Now, let's plug in the values:

$s = (24.5 \, \text{m/s}) (1.00 \, \text{s}) + \frac{1}{2} (-9.81 \, \text{m/s}^2) (1.00 \, \text{s})^2$

First, calculate each term separately:

1. $v_0 t = 24.5 \, \text{m}$
2. $\frac{1}{2} a t^2 = \frac{1}{2} \times (-9.81) \times (1.00)^2 = -4.905 \, \text{m}$

Now, sum them:

$s = 24.5 \, \text{m} - 4.905 \, \text{m} = 19.595 \, \text{m}$

So, the displacement of the rock after 1.00 second is 19.6 meters (rounded to one decimal place) upward.

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Do you want more details or have any questions?

Here are five related questions you might find useful:

1. What is the maximum height reached by the rock?
2. How long will it take for the rock to reach the highest point?
3. What will be the velocity of the rock after 2 seconds?
4. How long will the rock be in the air before it hits the ground?
5. How does air resistance affect the motion of the rock?

Tip: Always keep track of the signs for velocity and acceleration in kinematic equations to avoid mistakes.