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How will you derive the formula [v=√(2gh) ] ?

where is this formula derived from?
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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... Show More
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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 distancenow 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 fromso considering aal the above we havev^2 = 2gh or v=(2gh)^1/2ok i hope u have got u r ans and enjoy physics.............. Show More
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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 ).... Show More
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For the best answers, search on this site https://shorturl.im/avbZXA) 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.... Show More
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