Help with a mechanics problem from Kleppner's book

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SUMMARY

The discussion centers on a mechanics problem from Kleppner's book involving an elevator and a marble dropped by a boy. The key equations used are h = vT1 and h = vT2 - 0.5g(T2)^2, where g = 9.8 m/s². The user identifies a critical error in their calculations regarding the signage when T1 equals T2, which leads to an incorrect height of zero. The correct interpretation indicates that the height the elevator rises at T1 is equal to the height the marble falls at T2, resulting in a total distance of zero.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Familiarity with concepts of uniform velocity and constant acceleration
  • Knowledge of gravitational acceleration (g = 9.8 m/s²)
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Review kinematic equations for motion under constant acceleration
  • Study the concept of relative motion in physics
  • Explore problem-solving techniques for mechanics problems in introductory physics
  • Examine the implications of signage in physics equations
USEFUL FOR

Students studying physics, particularly those tackling mechanics problems, educators looking for teaching resources, and anyone interested in understanding the principles of motion and acceleration.

Tahmeed
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Homework Statement


An elevator departs the ground at t=0 With uniform velocity. At T1 a boy drops a marbel through the floor of the lift that falls under constant acceleration g=9.8 ms^-2. If it takes marbel T2 time to fall down, what was the height at the T1.

Homework Equations



h=vT1
h= vT2 -.5g(T2)^2

The Attempt at a Solution



I simply replaced v in second equation by h/T1. And i got the h in terms of T1, T2. But the problem is, there is a clue given with this problem in the book. That is when T1=T2=4 h=39.2 now, in my equation, T1=T2 gives a height 0.
 
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you have to relook your 2nd equation for proper signage.
Tahmeed said:

Homework Statement


An elevator departs the ground at t=0 With uniform velocity. At T1 a boy drops a marbel through the floor of the lift that falls under constant acceleration g=9.8 ms^-2. If it takes marbel T2 time to fall down, what was the height at the T1.

Homework Equations



h=vT1
h= vT2 -.5g(T2)^2

The Attempt at a Solution



I simply replaced v in second equation by h/T1. And i got the h in terms of T1, T2. But the problem is, there is a clue given with this problem in the book. That is when T1=T2=4 h=39.2 now, in my equation, T1=T2 gives a height 0.
the height that the elevator rises at T1 is the same as the height the marble falls at T2. The sum total of those 2 distances is 0 (h - h =0). Check signage.
 

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