Answer to Question #33840 in Mechanics | Relativity for DURGA
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.
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.