Question

A stone is dropped from the top of a tall cliff and 1sec. later a second stone is thrown vertically downward with a velocity of 20m/sec.How far below the top of the cliff will the second stone overtake the first?

Answer 1

Apply this formula and check if you are getting the answer or not?
et 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]
Put the values and you should get your answer!