Solving Physics Problems: Dropped Stones, Thrown Balls, and Falling Tiles

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SUMMARY

This discussion focuses on solving physics problems related to objects in free fall and projectile motion. The first problem involves calculating the depth of a hole based on the time it takes for a dropped stone to hit the bottom and the sound to travel back, using the speed of sound at 343 m/s. The second problem requires determining the average speed of a person running to catch a ball thrown from a 25.0-meter tall building with an initial speed of 12 m/s. The third problem calculates the height from which a roof tile falls, given the time it takes to pass a 1.8-meter window. All problems utilize principles of constant acceleration and kinematic equations.

PREREQUISITES
  • Understanding of kinematic equations for constant acceleration
  • Knowledge of the speed of sound in air (343 m/s)
  • Familiarity with projectile motion concepts
  • Basic algebra for solving equations
NEXT STEPS
  • Study the kinematic equations for free fall and projectile motion
  • Learn how to calculate time of flight for dropped objects
  • Explore the relationship between distance, speed, and time in motion problems
  • Practice solving real-world physics problems involving acceleration due to gravity
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in applying mathematical concepts to solve real-world motion problems.

SomeHustler
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1. A spelunker (cave explorer) drops a stone from rest into a hole. The speed of sound is 343 m/s in air, and the sound of the stone striking the bottom is heard 1.64 s after the stone is dropped. How deep is the hole?

2. A ball is thrown upward from the top of a 25.0-m-tall building. The ball's initial speed is 12 m/s. At the same instant, a person is running on the ground at a distance of 32.6 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

3. A roof tile falls from rest from the top of a building. An observer inside the building notices that it takes 0.18 s for the tile to pass her window, whose height is 1.8 m. How far above the top of this window is the roof?

Homework Equations


Acceleratior = (Vf + Vi) / time displacement

Vector average = (Vf + Vi) / 2

Vector Average = distance displacement / time displacement

The Attempt at a Solution


No clue, if someone could help me that would be great. I can't do either of those 3 for some reason.
 
Last edited:
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Give these a shot... all of them involve acceleration due to gravity. Have you studied equations for constant acceleration?
 

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