Question

a food packet is released from a helicopter which is rising steadily at 2 m/s after 2 sec ( i ) what is the velocity of the packet ( ii) how far is it below the helicopter?take g= 9.8m/s

Answer 1

Answer:

It is 19.6 m below the helicopter.

Explanation:

Given that,

Initial velocity = 2 m/s

Time t = 2 sec

(I). The velocity of the packet is

Using equation of motion

$v = u+gt$

$v = 2-9.8\times2$

$v =-17.6\ m/s$

Negative sign shows that the direction of downward.

(II)The distance covers by the helicopter is

$d = v\times t$

$d_{1} = 2\times 2$

$d_{1} = 4\ m$

The distance covers by the packet

Using equation of motion

$v^2=u^2-2gs$

$(17.6)^2=2^2-2\times9.8\times s$

$d_{2}=\dfrac{(17.6)^2-2^2}{2\times9.8}$

$d_{2} = 15.6$

The total distance is

$D=d_{1}+d_{2}$

$D=4+15.6$

$D=19.6\ m$

Hence, It is 19.6 m below the helicopter.

Answer 2

Explanation:

i hope that you will understand ☺️