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 .(Assume upward acceleration as -g and downward acceleration as +g)
Answers
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:
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