From point A, 80 m above …
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From point A, 80 m above the ground. A particle is projected vertically upwards with a velocity of 29.4m/s^2 . 5 secs. Later another particle is dropped from point B, 34.3 m vertically below from point A. Determine when and where one particle overtakes the other. ( use g=9.8m/s^2)
Posted by Diwakar Khandelwal 5 years, 4 months ago
- 1 answers
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Gaurav Seth 5 years, 4 months ago
Let at time t, they reach level C.
For A:- -(h + 34.3) = 29.4t - (1/2)gt2 ...(i)
For B:- -h = −21g(t−5)2 ...(ii)
(ii) - (i) ⇒ 34.3
= -29.4t + (1/2)g [t2−(t−5)2
⇒ 7 = -6t + (2t - 5) 5 ⇒ t = 8s
Now from (ii), h = (1/2) (9.8) (8−5)2 - 44.1 m.
From figure,
H = 80 - 34.3 = 45.7 m.
So finally, h1 = H - h = 45.7 - 44.1 = 1.6 m
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