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A ball is dropped from a height h above the ground. Neglect the air resistance, it's velocity v varies with its height y above the ground as a) √2g(h-y) b) √2gh c)√2gy d)√2g(h+y)
if height from which it has been dropped is `h` and for a point `y` m above the ground, the distance travelled by the ball will be (h-y). Answer - option AV² - U² = 2 * (-g)*(-(h-y))Since it has been dropped U = 0 and simplifying we get, V² = 2g(h-y)Hence V = √(2g(h-y))
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