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A stone projected vertically up from the ground reaches a height y in it's path at t1 seconds and after further t2 seconds reaches the ground. The height y is equal toA) (1/2)g(t1+t2)B) (1/2)g(t1+t2)2C) (1/2)g(t1×t2)D) g×t1×t2
Dear Ajith Let t1 be the time where particle reach height " h" and let t2 be the time where particle attains same height "h" again. Time of flight T=t1 +t2 h=(usinθt)+1/2gt2 ; u is velocity of projection. gt2-2usinθt+2h=0the roots of this equation are t1and t2t1t2=2h/g, t1+t2=2usinθ/g [for ax2+bx+c=0,sum of the roots=-b/a and product of the roots=c/a] By solving these equations we get h=1/2gt1t2
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