Two bodies 1 and 2 are projected simultaneously with velocities `v_1 = 2ms^(-1) and v_2 = 4ms^(-1)` respectively. The body 1 is projected vertically up from the top of a cliff of height h = 10m 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.

Answer

5

Explanation

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 = s_1, v_0 = v_1, a = -g`
Then, `s_1 = v_1t - 1/2 "gt"^2` .........(i)
For the second particle, `s = s_2, v_0 = v_1, a= -g`
Then , `s_2 = v_2t - 1/2 "gt"^2` ..........(ii)
Referring to the figure, `s_2 - s_1 = h`......(iii)
Substtituting `s_1` from Eq. (i), s_2 from Eq. (ii) in Eq. (iii), we have
` t= h/(v_2-v_1) = 10/(4-2) = 5s`.