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a body is projected vertically upwards . if t1 and t2 be the times at which the height h above the point of projection while ascending and descending respectively then (a)h=gt1t2 (b)h=2gt1t2 (c)h=1/2gt1t2 (d)h=1/4gt1t2

a body is projected vertically upwards . if t1 and t2 be the times at which the height h above the point of projection while ascending and descending respectively then 
(a)h=gt1t2
(b)h=2gt1t2
(c)h=1/2gt1t2
(d)h=1/4gt1t2

Grade:11

3 Answers

Arun
25750 Points
4 years ago
 
t1 is the time the particle takes to reach h.
so,
              h = ut1 - ½ gt12           \Rightarrow         u = (h + ½ gt12) / t1      (equation 1)
 
t2 is the time taken by the particle to reach the same point again.
so,
             h = ut2 - ½ gt22
 
   substituting 1 in the above equation:
 
            h = (h + ½ gt12) t2/t1 – ½ gt22
 
   solving
 
            h = ½ gt1t2
Khimraj
3007 Points
4 years ago
 
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=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
aswanth nayak
100 Points
4 years ago
Dear student,
 
 
 
 
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=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

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