Falling down motions under the sole influence of gravity
A stone is dropped from rest from a bridge which is H
= 63 m
above water. The magnitude of the gravitational acceleration is 9.8m/s2.

Take DOWNWARD as positive direction.

What is the stone's velocity at t = 2.0 seconds? Keep 2 digits after the decimal point.

To find the stone's velocity at \( t = 2.0 \) seconds, we can use the equation for velocity under constant acceleration. Since the stone is dropped from rest, its initial velocity (\(v_0\)) is 0.

The equation for velocity under constant acceleration is:

\[
v = v_0 + at
\]

Where:
- \( v \) is the final velocity,
- \( v_0 \) is the initial velocity,
- \( a \) is the acceleration (which is \(9.8 \, \text{m/s}^2\) in this case),
- \( t \) is the time.

Given:
- \( v_0 = 0 \, \text{m/s} \) (the stone is dropped, so the initial velocity is 0),
- \( a = 9.8 \, \text{m/s}^2 \),
- \( t = 2.0 \, \text{s} \).

Substitute these values into the equation:

\[
v = 0 + (9.8 \, \text{m/s}^2)(2.0 \, \text{s})
\]

\[
v = 19.6 \, \text{m/s}
\]

So, the stone's velocity at \( t = 2.0 \) seconds is **19.60 m/s**.

Would you like more details or have any questions?

Here are some related questions you might find interesting:
1. What would be the velocity of the stone at \( t = 5.0 \) seconds?
2. How long does it take for the stone to reach the water?
3. What is the velocity of the stone just before it hits the water?
4. How far does the stone fall in the first 3 seconds?
5. What would the velocity be if the stone was thrown downwards with an initial velocity?

**Tip:** When dealing with free fall problems, always pay attention to the sign convention for direction and initial conditions.

Explore how to calculate the velocity of a stone dropped from a height of 63m at t = 2.0 seconds due to gravitational acceleration. Learn the velocity formula under constant acceleration and solve for the stone's velocity step-by-step.

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