a food packet is released from a helicopter which is rising steadily at 3m/s. after two seconds how far is the food packet below the helicopter
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a food packet is released from a helicopter which is rising steadily at 3m/s. after two seconds how far is the food packet below the helicopter
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Answer:
.Initial velocity = 2 m/s
Time t = 2 sec
(I). The velocity of the packet is
Using equation of motion
v = u+gtv=u+gt
v = 2-9.8\times2v=2−9.8×2
v =-17.6\ m/sv=−17.6 m/s
Negative sign shows that the direction of downward.
(II)The distance covers by the helicopter is
d = v\times td=v×t
d_{1} = 2\times 2d1=2×2
d_{1} = 4\ md1=4 m
The distance covers by the packet
Using equation of motion
v^2=u^2-2gsv2=u2−2gs
(17.6)^2=2^2-2\times9.8\times s(17.6)2=22−2×9.8×s
d_{2}=\dfrac{(17.6)^2-2^2}{2\times9.8}d2=2×9.8(17.6)2−22
d_{2} = 15.6d2=15.6
The total distance is
D=d_{1}+d_{2}D=d1+d2
D=4+15.6D=4+15.6
D=19.6\ mD=19.6 m