Kinematics of thrown chalk - find time to hit the floor

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The discussion revolves around solving the kinematics problem of a chalk thrown upwards at 2 m/s from a height of 1 meter. Participants debate the requirement of using only one equation of motion to determine the time it takes for the chalk to hit the floor. One contributor suggests using the equation v = u + at, while another argues that a second equation is necessary for accurate calculation. The quadratic equation d = -(1/2)gt² + 2t + 1 is proposed, leading to two potential solutions for time, with the positive value being the relevant answer. The conversation highlights the challenge of adhering to the constraint of using a single equation for this scenario.
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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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