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)

Answer 1

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)`