To determine the velocity of the pea at each second, we can use the following kinematic equation for velocity under constant acceleration due to gravity:

$v = v_0 - g \cdot t$

Where:

• $v$ is the velocity at time $t$,
• $v_0$ is the initial velocity (30 m/s upwards),
• $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$ downwards),
• $t$ is the time in seconds.

Now let's calculate the velocity at each second:

1. At $t = 0$ sec: $v = 30 - 9.8 \times 0 = 30 \, \text{m/s}$ The velocity is $30 \, \text{m/s}$ (upwards).

2. At $t = 1$ sec: $v = 30 - 9.8 \times 1 = 30 - 9.8 = 20.2 \, \text{m/s}$ The velocity is $20.2 \, \text{m/s}$ (upwards).

3. At $t = 2$ sec: $v = 30 - 9.8 \times 2 = 30 - 19.6 = 10.4 \, \text{m/s}$ The velocity is $10.4 \, \text{m/s}$ (upwards).

4. At $t = 3$ sec: $v = 30 - 9.8 \times 3 = 30 - 29.4 = 0.6 \, \text{m/s}$ The velocity is $0.6 \, \text{m/s}$ (upwards, but very close to zero).

5. At $t = 4$ sec: $v = 30 - 9.8 \times 4 = 30 - 39.2 = -9.2 \, \text{m/s}$ The velocity is $-9.2 \, \text{m/s}$ (downwards).

6. At $t = 5$ sec: $v = 30 - 9.8 \times 5 = 30 - 49 = -19 \, \text{m/s}$ The velocity is $-19 \, \text{m/s}$ (downwards).

7. At $t = 6$ sec: $v = 30 - 9.8 \times 6 = 30 - 58.8 = -28.8 \, \text{m/s}$ The velocity is $-28.8 \, \text{m/s}$ (downwards).

Summary:

• $t = 0$ sec: $30 \, \text{m/s}$ (upwards),
• $t = 1$ sec: $20.2 \, \text{m/s}$ (upwards),
• $t = 2$ sec: $10.4 \, \text{m/s}$ (upwards),
• $t = 3$ sec: $0.6 \, \text{m/s}$ (upwards),
• $t = 4$ sec: $-9.2 \, \text{m/s}$ (downwards),
• $t = 5$ sec: $-19 \, \text{m/s}$ (downwards),
• $t = 6$ sec: $-28.8 \, \text{m/s}$ (downwards).

Would you like more details about any part of this, or do you have any questions?

Here are 5 related questions:

1. At what time does the pea reach its highest point?
2. How far does the pea travel upwards before coming to rest?
3. What is the total time the pea stays in the air?
4. What is the pea's acceleration at its highest point?
5. What will be the pea's velocity upon returning to the ground?

Tip: When an object is moving upward and downward under gravity, its velocity becomes zero at its peak, but acceleration remains constant at $-9.8 \, \text{m/s}^2$.