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 u1² :u2² .

Answer 1

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

Answer 2

Answer:

We know for upward motion

v2 = u2 – 2gh

h = (u2 – v2)/2g

At highest point v = 0

Therefore

h = u2/2g

For the first ball

h1 = u21/2g

For the second ball

h2 = u22/2g

Thus

h1/h2 = (u21/2g)/(u22/2g)

or

h1 : h2 = u22 : u21