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Question

Two particles $$1$$ and $$2$$ are thrown in the directions shown in figure simultaneously with velocities $$5 m/s$$ and $$20 m/s$$. Initially, particle $$1$$ is at height $$20 m$$ from the ground. Taking upwards as the positive direction, find the time when the particles will collide.

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Solution

Given :      $$u_1  = -5  m/s$$                 $$u_2  = 20  m/s$$
Let the particles collide at point P after time $$t$$
For particle 1 :        $$S_1  = u_1 t - \dfrac{1}{2} gt^2$$                  where  $$g =10  m/s^2$$
$$\therefore$$     $$-x   = -5t - 5 t^2  $$                             ..............(a)

For particle 2 :        $$S_2  = u_2 t - \dfrac{1}{2} gt^2$$            
$$\therefore$$     $$20-x   = 20t - 5 t^2  $$                             ..............(b)
Subtracting (a) from (b) we get        $$20   =25  t $$             $$\implies t  = 0.8$$  s

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Physics

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