To solve for the velocity and displacement of the rock at 4.0 seconds, we will use the following kinematic equations:

1. Velocity equation: $v_f = v_i + a \cdot t$ Where:

• $v_f$ = final velocity
• $v_i = -8.0 \, \text{m/s}$ (the negative sign indicates downward direction)
• $a = g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity, also in the downward direction)
• $t = 4.0 \, \text{seconds}$

2. Displacement equation: $d = v_i \cdot t + \frac{1}{2} a \cdot t^2$ Where:

• $d$ = displacement
• $v_i = -8.0 \, \text{m/s}$
• $a = 9.8 \, \text{m/s}^2$
• $t = 4.0 \, \text{seconds}$

Step 1: Finding the final velocity

Substitute the values into the velocity equation: $v_f = -8.0 \, \text{m/s} + (9.8 \, \text{m/s}^2 \cdot 4.0 \, \text{seconds})$ $v_f = -8.0 \, \text{m/s} + 39.2 \, \text{m/s} = 31.2 \, \text{m/s}$ So, the final velocity after 4 seconds is 31.2 m/s [down].

Step 2: Finding the displacement

Now, using the displacement equation: $d = (-8.0 \, \text{m/s}) \cdot 4.0 \, \text{seconds} + \frac{1}{2} (9.8 \, \text{m/s}^2) \cdot (4.0 \, \text{seconds})^2$ $d = -32.0 \, \text{m} + \frac{1}{2} \cdot 9.8 \, \text{m/s}^2 \cdot 16.0 \, \text{seconds}^2$ $d = -32.0 \, \text{m} + 78.4 \, \text{m}$ $d = 46.4 \, \text{m}$ So, the displacement after 4 seconds is 46.4 meters [down].

Final Answer:

• Final velocity: 31.2 m/s [down]
• Displacement: 46.4 m [down]

Would you like more details on any step, or help with another problem?

Here are five related questions to deepen your understanding:

1. What would the final velocity be if the initial velocity was zero?
2. How would the displacement change if the rock was thrown upward with 8.0 m/s?
3. What happens to the displacement if the time was reduced to 2 seconds?
4. How does the direction of velocity affect the motion equations?
5. What would be the time taken for the rock to hit the ground if thrown from 100 meters?

Tip: Always ensure that signs in kinematic equations match the direction of motion. This helps avoid confusion and incorrect answers.