Question

Two stones are thrown vertically upwards simultaneously with their initial velocities x m/s and y ms. Find
the ratio of maximum height reached by them.​

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}s=

2g

v

2

−u

2

For object 1:

s_1= \frac{v_1^2-u_1^2}{-2g}s

1

=

−2g

v

1

2

−u

1

2

Putting v=0

s_1= \frac{-u_1^2}{-2g}= \frac{u_1^2}{2g}s

1

=

−2g

−u

1

2

=

2g

u

1

2

..................(i)

Similarly for object 2:

s_2= \frac{u_2^2}{2g}s

2

=

2g

u

2

2

......................(ii)

Divide (i) by (ii)

\frac{s_1}{s_2} = \frac{ (\frac{u_1^2}{2g} )}{ (\frac{u_2^2}{2g}) }

s

2

s

1

=

(

2g

u

2

2

)

(

2g

u

1

2

)

Therefore,

\frac{s_1}{s_2}= \frac{u_1^2}{u_2^2} \ or \ u_1^2:u_2^2

s

2

s

1

=

u

2

2

u

1

2

or u

1

2

:u

2

2

Hence Proved