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 velocity of 1 with respect to 2 after time $$t= 0.5 s$$

Solution

## Given :      $$u_1 = -5 m/s$$                 $$u_2 = 20 m/s$$For particle 1 :        $$V_1 = u_1 - g t$$              where  $$t =0.5$$  s$$\therefore$$     $$V_1 = -5 - 10 \times 0.5 = -10 m/s$$For particle 2 :        $$V_2 = u_2 - g t$$              where  $$t =0.5$$  s$$\therefore$$     $$V_2 = 20 - 10 \times 0.5 = 15 m/s$$Relative velocity of 1 w.r.t 2 after  $$t = 0.5$$  s        $$V_{12} = V_1- V_2 = -10 - 15 = -25 m/s$$Physics

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