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 – ½gt^{2}

For height h, y = h

h = ut – ½gt^{2}

gt^{2} – 2ut + 2h = 0

Solving

=> t = {u – √(u^{2} – 2gh)}/g

Or

t = {u + √(u^{2} – 2gh)}/g

Here both roots are acceptable because the height h is observed for two times t = t_{1} and t = t_{1} + dt. First root height h at t = t_{1} and the second root gives the height h at t = t_{1} + dt

Let height observed for 1^{st} time at t = t_{1} s

So t_{1} = {u – √(u^{2} – 2gh)}/g ----1

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

t_{1} + dt = {u + √(u^{2} – 2gh)}/g ----2

by eqn 1 and 2

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

Simplifying,

=> dt = 2√(u^{2} – 2gh)/g

Squaring

=> (dt)^{2} = 4(u^{2} – 2gh)/g^{2}

(dt)^{2} = 0 because dt is small quantity

So,

u^{2} – 2gh = 0

=> u = √(2gh)

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