Two bodies 1 and 2 are projected simultaneously with velocities v1 = 2 ms-1 and v2 = 4 ms-1 respectively. The body 1 is projected vertically up from the top of a cliff of height h =  10 m and the body 2 is projected vertically up from the bottom of the cliff. If the bodies meet, find the time (in s) of meeting of the bodies.

Let the particle meet after a time t. First of all, we choose the point of collision above the top of the cliff.For the first particle, s=s1, v0=v1, a=−gThen, s1=v1t−12gt2          …(i)For the second particle, s=s2, v0=v2, a=−gThen, s2=v2t−12gt2         …(ii)Referring to the figure, s2−s1=h      …(iii)Substituting s1 from Eq. (i), s2 from Eq. (ii) in Eq. (iii), we havet=hv2−v1=104−2=5s