a ball is thrown vertically upwards.it was observed at a height h twice with a time interval dt the initial velocity of the ball is

Let the initial velocity be u

Equation of motion of the ball

y = ut – ½gt²

For height h, y = h 

h = ut – ½gt²  

gt² – 2ut + 2h = 0

Solving

=> t = {u – √(u² – 2gh)}/g

Or

t = {u + √(u² – 2gh)}/g

Here both roots are acceptable because the height h is observed for two times t = t₁­ and t = t₁ + dt. First root height h at t = t₁ and the second root gives the height h at t = t₁ + dt

Let height observed for 1^st time at  t = t₁ s

So t₁ = {u – √(u² – 2gh)}/g ----1

Again for the next time height h is observed after time dt. So,

t₁ + dt =  {u + √(u² – 2gh)}/g ----2

by eqn 1 and 2

dt + {u – √(u² – 2gh)}/g = {u + √(u² – 2gh)}/g

Simplifying,

=> dt =  2√(u² – 2gh)/g

Squaring

=> (dt)² = 4(u² – 2gh)/g²

(dt)² = 0 because dt is small quantity

So,

u² – 2gh = 0

=> u = √(2gh)