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
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
Similar questions