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 :

Question

Rahul G · Answer

taking +ve axis along downward direction of pillar , equation of first stone is equation of second is can be verified that if t=n then y=0 therefore stone is still at origin . now solving the two equations we get and substituting value of t in first we get which is the point of intersection of two stones in terms of given and also the point where stone two overtakes stone 1

Vipul Tiwari · Answer

let time taken by first stone be ts and that of the other stone be (t-n)s so,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 equal 1/2gt^2=u(t-n) + 1/2[g(t-n)^2] gt^2 = 2ut-2un + gt^2 + gn^2 -2gnt t(2gn-2u) = gn^2-2un t =(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]

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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 :

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 :

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