Solve Kinematic Problem: Mountain Climber Throws Stones

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SUMMARY

The discussion focuses on a kinematic problem involving a mountain climber who throws two stones from a 50.0m cliff. The first stone, thrown with an initial velocity of +2.0 m/s, takes 3.4 seconds to hit the water, while the second stone, thrown 1.0 second later, has an initial velocity of -9.06 m/s. Both stones hit the water simultaneously, and their final velocities can be calculated using kinematic equations. The climber's interpretation and calculations are confirmed as correct by other participants.

PREREQUISITES
  • Understanding of kinematic equations, specifically displacement and velocity formulas.
  • Knowledge of gravitational acceleration, specifically -9.81 m/s².
  • Familiarity with quadratic equations and their application in physics problems.
  • Ability to analyze motion in a vertical plane.
NEXT STEPS
  • Study the application of kinematic equations in free-fall scenarios.
  • Learn about the effects of initial velocity on projectile motion.
  • Explore the concept of simultaneous motion in physics problems.
  • Practice solving similar kinematic problems involving multiple objects.
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Students studying physics, educators teaching kinematics, and anyone interested in solving real-world motion problems involving gravity and projectile dynamics.

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Homework Statement



A mountain climber stands at the top of a 50.0m cliff hanging over a calm pool of water. The climber throws two stones vertically 1.0 s apart and observes that they cause a single splash when they hit the water. The first stone has an initial velocity of + 2.0 m/s.

a) How long after release of the first stone will the two stones hit the water?

b) What is the initial velocity of the second stone when it is thrown?

c) What will the velocity of each stone be at the instant both stones hit the water?

Homework Equations



displacement = V1time + 1/2atime2
V2 = V1 + a(time)

The Attempt at a Solution



For a, I'm essentially, interpreting it as the time is takes the first stone to hit the water, as that is the time after the release. And the second stone is thrown during this time. -50.0m for displacement, +2.0 for velocity 1, acceleration is -9.81. I rearrange and use the quad formula for solve to get the time as 3.4s. Is my interpretation of part a correct?

For b, I subtract 1.0s from the 3.4s, as the second is released 1.0s after. Same equation used in part a, to solve for V1, which turns out as -9.06 m/s. Number seems somewhat reasonable, since it falls much faster.

For c, I solve the final velocities separately, both using the formula for V2, but for differnet times and init. velocities.

Are my thought processes and method correct?
 
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