Question

A food packet is released from a helicopter which is rising steadily at 2m/s. the food packet falls on the ground after 6s. Find the height of the helicopter when (i) the food packet was released from it, and (ii) when the food packet just reached the Earth.

Answer 1

Answer:

It is 176.4 m below the helicopter.

Explanation:

Given that

Initial velocity = 2 m/s

Time t = 6 sec

(I). The velocity of the packet is

Using equation of motion

v = u + gt

v = 2 - 9.8 * 6

v = -56.8 m/s

Negative sign shows that the direction of downward.

(II)The distance covers by the helicopter is

d = vt

d₁ = 2 * 6

d₁ = 12 m

The distance covers by the packet

Using equation of motion

v² = u² - gs

(-56.8)² = 2² - 6*9.8 *s

d₂ = (56.8)² - 2²/ 2*9.8

d₂ = 164.4

The total distance is

D = d₁ + d₂

D = 12 + 164.4

⇒176.4

Hence, It is 176.4 m below the helicopter.

HOPE THAT HELPS :)