Monday, August 11, 2008

Q) A body is thrown vertically upwards from the top X of tower . It reaches the ground in A seconds . If it is thrown vertically downwards from X with the same speed, it reaches the ground in B seconds. If it is allowed to fall freely from X, the time taken by it to reach the ground is?

Let's consider the top of X as origin
S = ut + 1/2 at^2
-X = uA - gA^2 / 2 -----------------------(1)
(Here we have taken -X because this is the final displacement after going up a certain distance coming back then going to the bottom of the building.)
-X = -uB - gB^2/2
X = uB + gB^2/2 --------------------------(2)

third case:(free fall, initial velocity is zero)
X = gT^2/2 ------------------------(3)

Multiply (1) and (2) with B and A respectively, then solving, we get:
X(A+B) = (g/2)(AB^2 + A^2.B)
X = ABg/2
Substitute this in (3):
ABg/2 = gT^2/2
T^2 = AB
T = root (AB)


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