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.

Question

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.

Infinity Learn · Accepted Answer

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, $s_{1} = v_{1} t - \frac{1}{2} {gt}^{2}$ …(i)

For the second particle, $s = s_{2}, v_{0} = v_{2}, a = - g$

Then, $s_{2} = v_{2} t - \frac{1}{2} {gt}^{2}$ …(ii)

Referring to the figure, $s_{2} - s_{1} = h$ …(iii)

Substituting s₁ from Eq. (i), s₂ from Eq. (ii) in Eq. (iii), we have

$t = \frac{h}{v_{2} - v_{1}} = \frac{10}{4 - 2} = 5 s$

Rectilinear Motion

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