How Do You Calculate the Initial Drop Height in a Free Fall Problem?

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SUMMARY

The discussion centers on calculating the initial drop height of a stone in free fall, using the acceleration due to gravity at 9.8 m/s². The physics student measures the time it takes for the stone to fall past a 2.2 m tall window, which is 0.30 seconds. The average velocity of the stone during this interval is calculated to be approximately 7.33 m/s, while the velocity at the top of the window is determined to be about 9 m/s. To find the initial height above the window, the student needs to calculate the distance fallen using the kinematic equation.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Basic knowledge of free fall and gravitational acceleration
  • Ability to perform calculations involving average velocity
  • Familiarity with velocity vs. time graphs
NEXT STEPS
  • Study the kinematic equation: delta(x) = initial velocity * time + 0.5 * (a * t²)
  • Learn how to derive velocity from position-time graphs
  • Explore the concept of free fall and its implications in physics
  • Investigate the effects of air resistance on falling objects
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Physics students, educators, and anyone interested in understanding the principles of motion and free fall calculations.

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Homework Statement



A stone is dropped off the science building and accelerates, from rest, toward the ground at 9.8 m/s/s. A curious physics student looks out the third floor window as the stone falls past. She happens to have a stopwatch and she finds that it takes 0.30 sec for the stone to fall past the 2.2 m tall window. She then sketches the velocity vs. time plot shown below, but realizing she is late for lunch, she doesn't use the plot to analyze the motion of the stone.

a) What was the average velocity of the stone as it fell past the window?

b) What was the velocity of the stone at the top of the window?

c) From what height above the top of the window did the stone start its fall?

Homework Equations



delta(x)=initial velocity*time + .5(at^2)

The Attempt at a Solution



a) I got 2.2/.3 m/s

b) I used the equation above and got 8.99999 m/s

c) I was clueless with part C
 
Physics news on Phys.org
If a stone starts from rest, over what distance must it fall so that reaches a final speed of 2.2/3 m/s? That's the distance you re looking for.
 

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