Question

Two particles 1 and 2 are projected simultaneously with velocities $v_1$ and $v_2$, respectively. Particle 1 is projected vertically up from the top of a cliff of height h and particle 2 is projected vertically up from the bottom of the cliff. Find the time of meeting of particles, if the bodies meet (a) Above the top of the cliff, (b) Between the top and bottom of the cliff, and (c) below the bottom of the cliff, find the time of the meeting of the particles.

Answer 1

a. The point of collision above the top of the cliff. Let the particle meet after time t.

For the first particle,s = $s_1$, $v_0=v_1$, a = -g.

Then, $s_1=v_{1}t-\left(1/2\right)g{t}^2$ (i)

For the second particle,s = $s_2$, $v_0=v_2$, a = -g.

Then, $s_2=v_{2}t-\left(1/2\right)g{t}^2$ (ii)

Referring to Fig.

$s_2-s_1=h$

Substituting $s_1$ from (i), $s_2$ from (ii) in (iii), we have

$t=\frac{h}{(v_2-v_1}$

b. The point of collision of the particle in between the top and bottom of the cliff.

For the first particle,s = $-s_1$, $v_0=v_1$, a = -g.

Position-time relation for the first particle,

$-s_1=v_{1}t-\left(1/2\right)g{t}^2$

This gives $s_1=\left(1/2\right)g{t}^2-v_{1}t$ (i)

Similarly, for the second particle,s = $s_2$, $v_0=v_2$, a = -g.

Then $s_2=v_{2}t-\left(1/2\right)g{t}^2$ (ii)

Referring to the Fig., $s_1+s_2=h$ (iii)

Substituting $s_1$ from (i), $s_2$ from (ii) in (iii), we have

$t=\frac{h}{(v_2-v_1)}$

c. The point of collision of the particles below the bottom of the cliff (let us

assume a ditch at the base of the cliff).

For article 1, $s=-s_1,v_0=v_1$ and a = -g.

Position-time relation for the first particle

$-s_1=v_{1}t-\left(1/2\right)g{t}^2$

This gives $s_1=\left(1/2\right)g{t}^2-v_{1}t$ (i)

Similarly, for particle 2, $s=-s_1,v_0=v_1$ and a = -g.

We have $s_2=\left(1/2\right)g{t}^2-v_{2}t$ (ii)

From Fig. $s_1-s_2=h$ (iii)

Substituting $s_1$ from (i), $s_2$ from (ii) in (iii), we have

$t=\frac{h}{(v_2-v_1)}$