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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 to A) (1/2)g(t1+t2) B) (1/2)g(t1+t2) 2 C) (1/2)g(t1×t2) D) g×t1×t2

2 years ago

Answers : (1)

Arun
25285 Points
							
Dear Ajith
 
 
Let tbe 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=0
the roots of this equation are t1and t2
t1t2=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
2 years ago
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