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A ball is thrown vertically upwards it was observed twice at a height of 'h' and a time interval of 't'.calculate initial velocity of body in terms of 'g' 't' and 'h'
ANS=1/2root(g^2*t^2-8gh) .......?
if correct then please aproove ill post the solution
√take the up direction as +ve. here s=h.
substituting the value values in s=ut+1/2at2
h=ut-gt2
gt2-ut+h=0
t=u+√u2-4gh/2g and u-√u2-4gh/2g
now the time is given t...
u+√u2-4gh/2g - u-√u2-4gh/2g =t
solving this u2=g2t2+4gh
therefore u=√g2t2+4gh.
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