A  stone  is  dropped  into  the  water  from  a  bridge  44  m  above  the  water.  Another  stone  is  thrown vertically downward 1.0 s after the first stone was dropped. Both stones strike the water at the same time. What was the initial velocity of the second stone?

We can solve this problem using the equations of motion for both stones.

### Step 1: Equation for the first stone (dropped with initial velocity = 0)
The first stone is dropped from rest, meaning its initial velocity is \( v_1 = 0 \). The equation of motion for the first stone is:

\[
d = v_1 t + \frac{1}{2} g t^2
\]

where:
- \( d = 44 \, \text{m} \) (the height of the bridge),
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( v_1 = 0 \),
- \( t \) is the time it takes for the first stone to hit the water.

Substituting into the equation:

\[
44 = 0 + \frac{1}{2} \times 9.8 \times t^2
\]

Simplifying:

\[
44 = 4.9 t^2
\]

Solving for \( t \):

\[
t^2 = \frac{44}{4.9} \approx 8.98
\]

\[
t \approx \sqrt{8.98} \approx 3.0 \, \text{seconds}
\]

Thus, the first stone takes approximately 3.0 seconds to hit the water.

### Step 2: Equation for the second stone (thrown downward after 1 second)
The second stone is thrown downward 1 second after the first stone, so its time to hit the water is \( t_2 = 3.0 - 1.0 = 2.0 \, \text{seconds} \).

Let the initial velocity of the second stone be \( v_2 \). The equation of motion for the second stone is:

\[
d = v_2 t_2 + \frac{1}{2} g t_2^2
\]

Substituting the known values:

\[
44 = v_2 \times 2.0 + \frac{1}{2} \times 9.8 \times (2.0)^2
\]

Simplifying:

\[
44 = 2 v_2 + 19.6
\]

Solving for \( v_2 \):

\[
2 v_2 = 44 - 19.6
\]

\[
2 v_2 = 24.4
\]

\[
v_2 = \frac{24.4}{2} = 12.2 \, \text{m/s}
\]

### Final Answer:
The initial velocity of the second stone is \( 12.2 \, \text{m/s} \) downward.

Would you like further details or clarifications on this solution?

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Here are 5 related questions to expand your understanding:

1. What would happen if the second stone was thrown upward instead of downward?
2. How would the solution change if air resistance was considered?
3. What is the impact of different heights on the time difference between the two stones?
4. How can this problem be solved graphically?
5. How does the time of free fall change with different gravitational accelerations?

**Tip:** For any object in free fall, its velocity increases by approximately 9.8 m/s every second due to gravity (in the absence of air resistance).

A stone is dropped into the water from a bridge 44 m above the water. Another stone is thrown vertically downward 1.0 s after the first stone was dropped. Both stones strike the water at the same time. What was the initial velocity of the second stone?

Learn how to calculate the initial velocity of a stone thrown downward from a 44-meter-high bridge, using equations of motion and free fall principles. This problem provides a step-by-step solution suitable for students in Grades 9-12, involving key physics concepts such as kinematics and gravitational acceleration.

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