Kinematics of thrown chalk - find time to hit the floor

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Homework Help Overview

The discussion revolves around a kinematics problem involving a chalk thrown upwards by a teacher from a height of 1 meter with an initial velocity of 2 m/s. The objective is to determine the time it takes for the chalk to hit the floor using only one equation of motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster expresses difficulty in solving the problem using only one equation of motion, suggesting a reliance on multiple equations is common. Some participants propose using the equation v = u + at, while others suggest a quadratic approach involving height and time.

Discussion Status

Participants are exploring different interpretations of the problem and the requirements for using a single equation. Some guidance has been offered regarding the application of kinematic equations, but there is no explicit consensus on the approach to take.

Contextual Notes

The problem is constrained by the requirement to use only one equation of motion, which has led to differing opinions on how to approach the solution. There is also a discussion about the signs of the variables involved, particularly the direction of gravity.

jacyh
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Using only one equation of motion, how will you solve this:
Suppose a the tteacher's hand is 1m from the floor, and the teacher flicked a chalk upwards at 2m/s. How long will it take for the chalk to reach the floor?

I can't seem to use only one equation, it's always two equations or three. :cry:
 
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Its simple.If the teacher throws the chalk upwards with velocity 2 m/s , then on downward motion, while at the same point on its return motion , its velocity would be 2 m/s . Now apply v=u+at .

BJ
 
Dr.Brain said:
Its simple.If the teacher throws the chalk upwards with velocity 2 m/s , then on downward motion, while at the same point on its return motion , its velocity would be 2 m/s . Now apply v=u+at .

BJ

But to find the time at which it hit the floor you would have to use a second "equation of motion" and apparently this question requires only one.

I would vote for d= -(1/2)gt2+ 2t+ 1 where d is the height above the floor. Set d= 0 and solve for t.
 
Take h= -1m(downward), u = +2 m/s (upward) and "g" negative(downward) and solve eq h = ut + (1/2)gt^2.
it is quadratic in t gives two values of t, the positive value is the answer.
 

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