Question

# Two particles 1 and 2 are projected simultaneously with velocities v1 and v2, 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.

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Solution

## a. The point of collision above the top of the cliff. Let the particle meet after time t.For the first particle,s = s1, v0=v1, a = -g.Then, s1=v1t−(1/2)gt2 (i)For the second particle,s = s2, v0=v2, a = -g.Then, s2=v2t−(1/2)gt2 (ii)Referring to Fig.s2−s1=hSubstituting s1 from (i), s2 from (ii) in (iii), we havet=h(v2−v1b. The point of collision of the particle in between the top and bottom of the cliff.For the first particle,s = −s1, v0=v1, a = -g.Position-time relation for the first particle,−s1=v1t−(1/2)gt2This gives s1=(1/2)gt2−v1t (i)Similarly, for the second particle,s = s2, v0=v2, a = -g.Then s2=v2t−(1/2)gt2 (ii)Referring to the Fig., s1+s2=h (iii)Substituting s1 from (i), s2 from (ii) in (iii), we havet=h(v2−v1)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=−s1,v0=v1 and a = -g.Position-time relation for the first particle−s1=v1t−(1/2)gt2This gives s1=(1/2)gt2−v1t (i)Similarly, for particle 2, s=−s1,v0=v1 and a = -g.We have s2=(1/2)gt2−v2t (ii)From Fig. s1−s2=h (iii)Substituting s1 from (i), s2 from (ii) in (iii), we havet=h(v2−v1)

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