Falling Particle: Solving for Initial Velocity and Acceleration

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SUMMARY

The discussion focuses on calculating the initial velocity (v0) and acceleration (a) of a particle thrown directly upwards, which passes a distance "d" twice—once while ascending at time "t1" and again while descending at time "t2". The established formulas for these calculations are v0 = d * [(t1 + t2) / (t1 * t2)] and a = (2 * d) / (t1 * t2). The problem is framed within the context of motion under constant acceleration, with the initial height set to zero.

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Homework Statement



Particle was thrown directly upwards. Point which is in distance “d” from the beginning of movement is passed by the particle two times. Once going upwards in time “t1” and second time going downwards in time “t2” from the beginning of the movement. What was the initial speed(v0) and the acceleration(a) of the particle?

Homework Equations



I really do not know how to start with this example.

The Attempt at a Solution



I have tried every possible solution that I was capable of, but I’m not able to come up with the correct answer which should be

v0=d*[(t1+t2)/(t1*t2)]

and

a=(2*d)/(t1*t2)
 
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Start with the equation for position o an object under constant acceleration.

Point which is in distance “d” from the beginning of movement is passed by the particle two times.

This defines the coordinate system for you: initial velocity is v0 and initial height is 0.
 

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