Question

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

Answer 1

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