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

244166_16d1dc8efa4b40c58b2f63e570951d57.png


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        $$V_{12}  = V_1- V_2  = -10 - 15  = -25  m/s$$

517696_244166_ans_6fd7098609984cfbb7074cdf9db5ca44.png

Physics

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