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Question
Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of `"u"_1^2 : "u"_2^2` (Assume upward acceleration is –g and downward acceleration to be +g)
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Solution
At the highest point, v = 0
For the sone thrown with velocity u1
`0 - "u"_1^2 = 2(-"g')h_1` (Using v2 − u2 = 2s)
or `"h"_1 = ("u"_1^2)/(2"g")`
Similarly for the stone thrown with velocity u2,
`"h"_2 = ("u"_2^2)/(2"g")`
∴ The required ratio, `"h"_1/"h"_2 = ("u"_1^2)/(2"g") xx (2"g")/("u"_2^2) = ("u"_1^2)/("u"_2^2)`
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