# 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 :

Rahul G
17 Points
6 years ago
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
Vipul Tiwari
21 Points
6 years ago
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]