(1/2) m v^2 = m g H
so
v = sqrt(2 g H)
the second time Vg = speed at ground
final Ke = initial Ke + amount gained in fall
(1/2) m Vg^2 = (1/2)mv^2 + mgh
(1/2) m Vg^2 = (1/2)mv^2 + (1/2)mv^2
Vg^2 = 2 v^2
Vg = v sqrt 2
An object is dropped from a height H and reaches the ground with a velocity v. The same object is thrown down from the same height H with an initial velocity v and reaches the ground with a velocity v2. Which of the following is the relationship between v2 and v?
The answer is v2=√(2)v. Can someone explain why?
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