Question

When an object is dropped from a height, its initial velocity, u = 0.
So, from the third equation of motion, v² – u² = 2gh
=> v² – 0² = 2gh
=> v² = 2gh.... equation (1).

When an object is dropped from a height, its mechanical energy before dropping, ME¹ = KE¹ + PE¹
=> ME¹ = 0 + mgh
=> ME¹ = mgh.
Again, when the object is dropped, its mechanical energy before touching the ground, ME² = KE² + PE ²
=> ME² = ½mv² + 0
=> ME² = ½mv².
According to law of conservation of energy,
ME¹ = ME²
=> mgh = ½mv²
=> gh = ½v²
=> 2gh = v²
=> v² = √(2gh).... equation (2).

So, equation (1) states that v = 2gh and equation (2) states that v = √(2gh).
But, equation (1) contradicts equation (2).
What's the correct value of v then?
Is v = 2gh, or is v =√(2gh) ???​

Answer 1

Answer:

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