Question

Q1. Two stones are thrown vertically upwards simultaneously with their initial velocities uj and u2 respectively. Prove that the heights reached by them would be in the ratio of u : u5. (Assume upward acceleration is -g and downward acceleration to be +g).


Please koi mere help kardo (╥﹏╥)​

Answer 1

Step-by-step explanation:

We know for upward motion, v2 = u2 – 2 g h or h = u2 -v2 /2g

We know for upward motion, v2 = u2 – 2 g h or h = u2 -v2 /2g But at highest point v = 0

We know for upward motion, v2 = u2 – 2 g h or h = u2 -v2 /2g But at highest point v = 0Therefore, h = u2 /2g

We know for upward motion, v2 = u2 – 2 g h or h = u2 -v2 /2g But at highest point v = 0Therefore, h = u2 /2g For first ball, h1 = u2 / 2g

We know for upward motion, v2 = u2 – 2 g h or h = u2 -v2 /2g But at highest point v = 0Therefore, h = u2 /2g For first ball, h1 = u2 / 2g and for second ball, h2= u2 / 2g

Answer 2

$answer$

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