Question 10

10.
  ​Two particles are projected simultaneously at t = 0 from the top of a tower from a common projection point with a horizontal projection velocity of 24 m/s and a vertical projection velocity of 32 m/s respectively. Then after 2 secs, the distance between the two will be ?

(1)  112 m                                                     (2)  80 m
(3)  40 m                                                       (4)  20 m

position of the particle projected 32m/sy=32*2-10*42=44 m 0,44position of the particle projected 24m/sx=24*2=48 my=-10*42=-20 m coordinate of this particel =48,-20d=482+642=80 m Regards

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