two stones are thrown vertically upwards simultaneously.Take their initial velocities as U1 AND U2.Prove that the heights reached by them(distance) is in the ratio of (u1)2 : (u2)2.(Assume the upward acceleration to be (-g) and downward acceleartion to be (+g).
pls answer step by step
Verified answer
use 3rd kinematic equation:-
vf^2=vi^2+2as
for first stone-(assume height reached = s1)
0^2=u1^2+2*-g*s1(as after reaching certain height it will stop)
-u1^2=-2gs1
s1=-u1^2/-2g...(1)
similarly for second stone we have height:-(assume height reached = s2)
s2=-u2^2/-2g...(2)
divide eq 1 and 2 we get:-
ratio=u1^2/u2^2..ans!
u1 / u2 = 1 where u1 = u2 = (2gs)^1/2