Determining the time it takes to reach half of a penny's terminal speed

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SUMMARY

The terminal speed of a penny is established at 11 m/s. To determine the time it takes for the penny to reach half of this speed (5.5 m/s) from rest, the appropriate kinematic equation is required. The equation Vf - Vi = a * delta T can be simplified by recognizing that the penny's acceleration is due to gravity (approximately 9.81 m/s²). The correct approach involves calculating the time using the formula delta T = (Vf - Vi) / a, where Vf is 5.5 m/s and Vi is 0 m/s.

PREREQUISITES
  • Understanding of kinematic equations
  • Basic knowledge of acceleration due to gravity (9.81 m/s²)
  • Familiarity with terminal velocity concepts
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Review kinematic equations for uniformly accelerated motion
  • Study the concept of terminal velocity in fluid dynamics
  • Practice solving problems involving free fall and air resistance
  • Explore the effects of different masses on terminal velocity
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Students studying physics, educators teaching mechanics, and anyone interested in understanding motion under gravity without air resistance.

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The question is:
The terminal speed of a penny is 11 m/s. By neglecting air resistance, calculate how long it takes for a penny falling from rest to reach half of this speed


Homework Equations


I thought maybe you had to use one of the kinematic equations but that didnt work


The Attempt at a Solution


I tried Vf-Vi/a = delta T
Also, I tried dividing 11 m/s by gravity and that didnt work.
Im not sure if I am totally missing an equation needed or if I am making this way harder than it really is. Any help would be greatly appreciated!
 
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you are on track with the answer, but why use 11ms-1? (as you stated earlier that it was the terminal velocity, and you were going to use half of the terminal velocity.

Also, the equation does not need to be so complicated. Look at these ones, and figure out which one is needed, by knowing the data you have and the what is in the equations.
 

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