A mountain climber stands at the top of a 35.0 m cliff that overhangs a calm pool of water. He throws two stones vertically downward 1.0 s apart and observes that they cause a single splash. The first stone has an initial velocity of -2.40 m/s.
(a) How long after release of the first stone will the two stones hit the water?
2.9326
 
Incorrect: Your answer is incorrect.
 s
(b) What initial velocity must the second stone have if they are to hit simultaneously?
27.8
 
Incorrect: Your answer is incorrect.
 m/s
(c) What will the velocity of each stone be at the instant they hit the water?

Let's carefully go through each part of the problem step by step.

### Given Information:
- Height of the cliff: \( h = 35.0 \, \text{m} \)
- Initial velocity of the first stone: \( v_{1i} = -2.40 \, \text{m/s} \)
- Time difference between releasing the two stones: \( \Delta t = 1.0 \, \text{s} \)
- Acceleration due to gravity: \( g = 9.81 \, \text{m/s}^2 \)

We need to find:
(a) The time after the first stone is released when the two stones hit the water.
(b) The initial velocity of the second stone so that both stones hit the water simultaneously.
(c) The velocity of each stone at the moment they hit the water.

---

### Part (a): Time after release when both stones hit the water

For the first stone, we use the equation of motion:

\[
h = v_{1i} t + \frac{1}{2} g t^2
\]

Plugging in the known values:

\[
35.0 = (-2.40) t + \frac{1}{2} (9.81) t^2
\]

This simplifies to:

\[
35.0 = -2.40t + 4.905t^2
\]

This is a quadratic equation of the form:

\[
4.905t^2 - 2.40t - 35.0 = 0
\]

We can now solve for \( t \) using the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 4.905 \), \( b = -2.40 \), and \( c = -35.0 \). Let's calculate the solution.

Solving this equation will give us the time \( t \) after the release of the first stone when both stones hit the water.The positive solution to the quadratic equation is:

\[
t = 2.93 \, \text{seconds}
\]

Thus, the first stone will hit the water 2.93 seconds after it is released.

---

### Part (b): Initial velocity of the second stone

The second stone is released 1.0 second after the first stone, so it has \( t - 1.0 = 2.93 - 1.0 = 1.93 \, \text{seconds} \) to reach the water.

We use the same equation of motion for the second stone but solve for its initial velocity \( v_{2i} \):

\[
h = v_{2i} t_2 + \frac{1}{2} g t_2^2
\]

where \( t_2 = 1.93 \, \text{seconds} \). Plugging in the values:

\[
35.0 = v_{2i} (1.93) + \frac{1}{2} (9.81) (1.93)^2
\]

This simplifies to:

\[
35.0 = 1.93 v_{2i} + 18.23
\]

Now, solving for \( v_{2i} \).It seems like I can’t do more advanced data analysis right now. Please try again later. Let me know if you'd like me to walk you through the calculations manually or assist with other questions!

A mountain climber stands at the top of a 35.0 m cliff that overhangs a calm pool of water. He throws two stones vertically downward 1.0 s apart and observes that they cause a single splash. The first stone has an initial velocity of -2.40 m/s.
(a) How long after release of the first stone will the two stones hit the water?
(b) What initial velocity must the second stone have if they are to hit simultaneously?
(c) What will the velocity of each stone be at the instant they hit the water?

A detailed step-by-step solution to a kinematics problem where a mountain climber throws two stones from a 35.0 m cliff, and both stones hit the water simultaneously. Learn how to calculate the time, initial velocity, and final velocity using equations of motion and the quadratic formula. Suitable for Grades 11-12 students.

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