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.

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        $$20 =25 t$$             $$\implies t = 0.8$$  sPhysics

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