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indrayani bargaje Grade: 10
        2 particles are thrown upwards along the same line with same initial velocity u=30m/s but at different times. the second particle is thrown T seconds after the first. what should be the minimum value of T so that the particles collide in mid air?? 
6 years ago

Answers : (1)

Sudheesh Singanamalla
114 Points
										

for the first particle final velocity 0 is achieved in ..


v = u - gt


t = -30 / -10 = 3 seconds ------------------------------------------ (i)


distance travelled


S = ut - gt^2 / 2


S = 90 - 90/2


S = 90 - 45 =  45 m


assume that the second body is thrown after 'x' seconds.


it has to reach the top at the same time as that of the 1st particle because of the velocity but at some point they will collide.


they collide at a point when the first particle is returning.


distance travelled in 4 seconds = 45 + ut + at^2 / 2


                                            = 45 + 0 + 5 = 50m but the displacement is 50-10 = 40m from the ground


distance travelled by second body in 2 seconds = ut - at^2/2


                                                                   = 60 - 10*4/2


                                                                   = 60 - 20 = 40m


so the two particles collide at a height of 40m when the first body has travelled for 4 seconds while the second body has travelled for 2 seconds, so the time difference gives the 'T' (time after the throw of 1st body)


T = 4-2 = 2 seconds


so the second body was thrown 2 seconds after the first body was thrown.


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PLEASE APPROVE


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6 years ago
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