To solve this problem, we need to find the velocity of the object at different times, as well as the velocity when it strikes the ground.

Step 1: Find the velocity function.

The velocity of the object is the derivative of the position function $s(t) = 300 - 4.9t^2$. The velocity function $v(t)$ is given by:

[ v(t) = \frac{d}{dt}[s(t)] = \frac{d}{dt}[300 - 4.9t^2] ]

Taking the derivative:

$v(t) = -9.8t$

This is the velocity function for the object.

Step 2: Calculate the velocity at $t = 2$, $t = 3$, and $t = 4$ seconds.

• For $t = 2$:

$v(2) = -9.8(2) = -19.6 \, \text{m/s}$

• For $t = 3$:

$v(3) = -9.8(3) = -29.4 \, \text{m/s}$

• For $t = 4$:

$v(4) = -9.8(4) = -39.2 \, \text{m/s}$

Step 3: Find the time when the object strikes the ground.

The object strikes the ground when $s(t) = 0$, which gives us:

$300 - 4.9t^2 = 0$

Solving for $t$:

$4.9t^2 = 300$ $t^2 = \frac{300}{4.9} \approx 61.22$ $t \approx \sqrt{61.22} \approx 7.83 \, \text{seconds}$

Step 4: Calculate the velocity when the object strikes the ground.

The velocity at $t \approx 7.83$ seconds is:

$v(7.83) = -9.8(7.83) \approx -76.73 \, \text{m/s}$

Thus, the velocity when the object strikes the ground is approximately $-76.73 \, \text{m/s}$.

Final Results:

• Velocity at $t = 2$ seconds: $-19.6 \, \text{m/s}$
• Velocity at $t = 3$ seconds: $-29.4 \, \text{m/s}$
• Velocity at $t = 4$ seconds: $-39.2 \, \text{m/s}$
• Velocity when the object strikes the ground: $-76.73 \, \text{m/s}$ (rounded to two decimal places).

Would you like further details or have any questions?

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Here are some follow-up questions you might want to explore:

1. What is the position of the object at $t = 2.5$ seconds?
2. How long does it take for the object to fall halfway to the ground?
3. What is the object's acceleration at any point during its fall?
4. How does air resistance affect the velocity of a real object in free fall?
5. What would the velocity be at $t = 5$ seconds?

Tip: The velocity is negative because the object is moving downward.