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Question

Two particles 1 and 2 are thrown in the directions shown in figure simultaneously with velocities 5m/s and 20m/s. Initially, particle 1 is at height 20m from the ground. Taking upwards as the positive direction, the velocity of 2 with respect to 1 is (20+x)m/s. Find the value of x.
244164_c34fdcfc5b964537a3c4c401aca3a1df.png

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Solution

The velocity of particle 1 at any time t is given as,
v1=u1gt=5gt
The velocity of particle 2 at any time t is given as,
v2=u2gt=20gt
Thus, relative velocity of particle 2 with respect to particle 1 is,
v21=v2v1=25m/s
x+20=25x=5

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