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A stone is dropped from the top of a tall cliff and n seconds later another stone is thrown vertically downwards with velocity u. Then the second stone overtakes the first, below the top of cliff at a distance given by :


3 years ago

## Answers : (2)

Rahul G
17 Points
							taking +ve axis along downward direction of pillar , equation of first stone is $y= \frac{1}{2}gt^{2}$                                                                                 equation of second is  $y=u(t-n)+ \frac{1}{2}g(t-n)^{2}$ can be verified that if t=n then y=0 therefore stone is still at origin .now solving the two equations we get $t=\frac{ng}{u}[1-\frac{n}{2}]+n$and substituting value of t in first we get $y=\frac{1}{2}g[\frac{ng}{u}(1-\frac{n}{2})+n]^{2}$ which is the point of intersection of two stones in terms of given and also the point where stone two overtakes stone 1

3 years ago
Vipul Tiwari
21 Points
							let time taken by first stone be ts and that of the other stone be (t-n)sso,distance travelled by first stone is :s=1/2(gt^2)..........eq^n(1)and that of the other stone is :s=u(t-n) + 1/2[g(t-n)^2].........eq^n(2)since both the stones meet at the distance so eq^n(1) and eq^n(2) will be equal1/2gt^2=u(t-n) + 1/2[g(t-n)^2]gt^2 = 2ut-2un + gt^2 + gn^2 -2gntt(2gn-2u) = gn^2-2unt =(gn^2-2un)/(2gn-2u)t=n(gn/2 – u)/(gn-u)now putting value of t in eq^n(1)s=1/2[g{n(gn/2 – u)/(gn-u)}^2]s=g/2[n(gn/2 – u)/(gn-u)^2]

3 years ago
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