How will you derive the formula [v=√(2gh) ] ?
Could you please explain how the formula ( v = \sqrt{2gh} ) is derived? What are the underlying principles or concepts that lead to this equation, and in what contexts is it typically applied?
5 Answers
mgh = 1/2 mv^2. that is, potential energy = kinetic energy. now, multiply both sides by 2 , that gives you 2mgh=mv^2. now divide both sides by m, that gives you 2gh=v^2. therefore √2gh = v
o man u i hope u know the formula v^2 - u ^2 = 2as
where av final vel. u is intial vel. a is acc. and s is distance
now when we drop an object then its initial vel. u = 0 and the acceleration a it experience is = g gravitational acceleration and s is the distance it travels that is the height it is thrown from
so considering aal the above we have
v^2 = 2gh or v=(2gh)^1/2
ok i hope u have got u r ans and enjoy physics...........
if a body falls freely under gravity from a height 'h', then using the relation V^2 - U^2 = 2 a S, where U = 0 and a = g , S = h , we can have V^2 - 0 = 2 g h from which you can see that
V = square root of ( 2 g h ).Dec 09, 2024
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A) PE = KE → m*g*h = ½*m*v² → v = √(2gh) B) a = constant = g; v = ∫a*dt = g*t; h = ∫v*dt = ½g*t² C) Solve (B) for t.
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