Question #33840

a free falling body crosses point P,Q and R with velocities V,2V AND 3V.find the ratio of the distance PQ to QR.

Expert's answer

PQ:

On PQ the velocity changes by the law:

2V = V + gt1

Then we get the time required to move from P to Q:

t1 = V / g

Law for changing coordinate:

PQ = V * t1 + g(t1)^2 / 2 = V^2 / g + g * (V/g)^2 / 2 = V^2 / g + V^2 / 2g =

3V^2 / 2g

So PQ = 3V^2 / 2g

QR:

On PQ the velocity changes by the law:

3V = 2V + gt2

Then we get the time from Q to R:

t2 = V / g

QR = 2V * t2 + g(t2)^2 / 2 = 2V^2 / g + g * (V/g)^2 / 2 = 5V^2 / 2g

So QR = 5V^2 / 2g

Finally the ratio is: PQ / QR = (3V^2 / 2g) / (5V^2 / 2g) = 3 / 5 = 0.6.

On PQ the velocity changes by the law:

2V = V + gt1

Then we get the time required to move from P to Q:

t1 = V / g

Law for changing coordinate:

PQ = V * t1 + g(t1)^2 / 2 = V^2 / g + g * (V/g)^2 / 2 = V^2 / g + V^2 / 2g =

3V^2 / 2g

So PQ = 3V^2 / 2g

QR:

On PQ the velocity changes by the law:

3V = 2V + gt2

Then we get the time from Q to R:

t2 = V / g

QR = 2V * t2 + g(t2)^2 / 2 = 2V^2 / g + g * (V/g)^2 / 2 = 5V^2 / 2g

So QR = 5V^2 / 2g

Finally the ratio is: PQ / QR = (3V^2 / 2g) / (5V^2 / 2g) = 3 / 5 = 0.6.

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