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 1st 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)

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