How long (in seconds) does it take the stone to hit the water? Keep 2 digits after the decimal point.

To determine how long it takes for the stone to hit the water, we can use the equation of motion for distance traveled under constant acceleration. The distance traveled \(s\) under constant acceleration \(a\) when starting from rest is given by:

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

Given:
- \( s = H = 63 \, \text{m} \) (height of the bridge),
- \( v_0 = 0 \, \text{m/s} \) (initial velocity),
- \( a = 9.8 \, \text{m/s}^2 \) (gravitational acceleration).

Since the stone is dropped from rest, the equation simplifies to:

\[
63 = 0 + \frac{1}{2} (9.8) t^2
\]

Simplify and solve for \(t\):

\[
63 = 4.9t^2
\]

\[
t^2 = \frac{63}{4.9}
\]

\[
t^2 = 12.8571
\]

\[
t = \sqrt{12.8571}
\]

\[
t \approx 3.58 \, \text{seconds}
\]

So, it takes approximately **3.58 seconds** for the stone to hit the water.

Would you like more details or have any questions?

Here are some related questions you might find interesting:
1. What is the final velocity of the stone when it hits the water?
2. How far does the stone fall in the first 2 seconds?
3. How much time does it take to fall the first half of the distance to the water?
4. If the stone was thrown downwards with an initial velocity, how would the time to hit the water change?
5. What would be the total distance traveled if the stone bounced back up after hitting the water?

**Tip:** For problems involving free fall, the square root function often appears when solving for time because of the \( t^2 \) term in the equations of motion.

Learn how to calculate the time it takes for a stone to hit water when dropped from a bridge. Solve using equations of motion and gravitational acceleration. Suitable for high school physics students.

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