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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`
5 months ago

```							 t1 is the time the particle takes to reach h.so,              h = ut1 - ½ gt12                    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
```
5 months 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=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
```
4 months 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=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
```
3 months ago
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