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 U1:U2.(assume upward acceleration is-g and downward acceleration is +g). plz don't spam I'm giving you 50 points plz don't spam and step by step explanation I will mark brainliest​

Answer 1

Answer:

for the first stone thrown with initial velocity u₁ ,

u₁ = initial velocity

v₁ = final velocity at maximum height = 0 m/s

a = acceleration = - g

h₁ = maximum height gained

using the equation

v²₁ = u²₁ + 2 a h₁

0² = u²₁ + 2 (- g)h₁

h₁ = u²₁ /(2g)                             eq-1

for the second stone thrown with initial velocity u₂ ,

u₂ = initial velocity

v₂ = final velocity at maximum height = 0 m/s

a = acceleration = - g

h₂ = maximum height gained

using the equation

v²₂ = u²₂ + 2 a h₂

0² = u²₂ + 2 (- g)h₂

h₂ = u²₂ /(2g)                             eq-2

dividing eq-1 by eq-2

h₁/h₂ = (u²₁ /(2g) )/(u²₂ /(2g))

h₁/h₂ = u²₁ /u²₂

Answer 2

$\huge \colorbox{red}{ꪖꪀᦓ᭙ꫀ᥅}$

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 :-

$\bold \orange{s= \frac{v^2-u^2}{2g}s=2gv2−u2 }$

For object 1:

$\bold \green{s_1= \frac{v_1^2-u_1^2}{-2g}s1=−2gv12−u12 }</p><p>$

$\fbox \red{ </p><p>Putting v=0}$

$\bold \blue{s_1= \frac{-u_1^2}{-2g}= \frac{u_1^2}{2g}s1=−2g−u12=2gu12}$

 (i)Similarly for object 2:-

$\bold \pink{s_2= \frac{u_2^2}{2g}s2=2 \: gu \: 22 .}$

❤Divide (i) by (ii)

$\bold \red{\frac{s_1}{s_2} = \frac{ (\frac{u_1^2}{2g} )}{ (\frac{u_2^2}{2g}) }s2s1=(2gu22)(2gu12) }$

Therefore :-

$\huge \pmb{\frac{s_1}{s_2}= \frac{u_1^2}{u_2^2} \ or \ u_1^2:u_2^2s2s1=u \: 22 \: u12 or u12:u22 }$

Hence Proved!~❤