Answers

2015-06-13T20:42:42+05:30
Given, initial velocities: u₁ and u₂
g=acceleration due to gravity, here it is (-g), as the object is thrown vertically upwards
Let Final velocities: v₁ and v₂
Now, we know that, 
s= \frac{v^2-u^2}{2g}
For object 1:
s_1= \frac{v_1^2-u_1^2}{-2g}
Putting v=0
s_1= \frac{-u_1^2}{-2g}= \frac{u_1^2}{2g}  ..................(i)
Similarly for object 2:
s_2= \frac{u_2^2}{2g} ......................(ii)
Divide (i) by (ii)
 \frac{s_1}{s_2} = \frac{ (\frac{u_1^2}{2g} )}{ (\frac{u_2^2}{2g}) }
Therefore,
 \frac{s_1}{s_2}= \frac{u_1^2}{u_2^2} \ or \ u_1^2:u_2^2
Hence Proved



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