Physics, asked by prateekyadav86, 10 months ago

If v be the instantaneous velocity of the body dropped from the top of a tower, when it is located at height h, then which of the following remains constant
(a) gh+ v^2/2
(b) gh+v^2
(d) gh - v^2
(c) gh-v^2/2​

Answers

Answered by smanavparashar20
1

Answer:

The correct answer is (a) gh+v^2/2

Kinetic energy + potential energy = constant

= 1/2mv^2 + mgh ( m cancels out both sides )

= 1/2v^2 + gh

= gh + v^2/2

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