For a body projected from top of a tower of height h, with horizontal velocity u, the time of flight is
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a
2hg
b
hg
c
g2h
d
gh
answer is A.
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Detailed Solution
for horizontal projected body let initial velocity =u;angle of projection =0 horizontal component of u is=ux=ucos0=u vertical component of u is=uy=usin0=0 time of flight t is estimated in vertical direction , applying sy=uyt+12gt2 ; here g=acceleration due to gravity here sy=h=height of the tower, hence on substituting h=0+12gt2 is obtained ⇒time of flight=2hg