Asked by khushal305 | 13th Jul, 2010, 04:19: PM
Given, vh (velocity of helicopter) = 2 m/s.
u = 0
Lets take g = 10 m/s2
(i) Using v = u + at we get,
v = 0 + 10x2
= 20 m/s.
Therefore velocity of packet after 2 seconds is 20m/s.
(ii) As the helicopter and packet are moving in opposite direction the total distance betweent the two will be the sum of the distances travelled by both in two seconds.
Distance travelled by object - S1 = ut + ½ at2
= 0 + ½ (10) (4)
= 20 m.
Distance travelled by helicopter - S2 = u't + ½ a't2
= 2x2 + 0
= 4 m.
Therefore distance between helicopter and the object after 2 seconds = 20 + 4 = 24 m.
Answered by | 16th Jul, 2010, 05:14: PM
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