Aswallow is carrying a coconut while flying at 28 m/s east when it loses its grip in the coconut falls if the swallows flying 50 m above the ground how far does the coconut land in front of where it was dropped

89·432 m

Explanation:

Initial velocity of the coconut = 28 m/s

Initial velocity of the coconut will be the initial horizontal velocity of the coconut

As there is no horizontal acceleration, the horizontal velocity of the coconut remains constant

Therefore at any instant the horizontal velocity of the coconut = 28 m/s

Initial height of the coconut above the ground = 50 m

Let the time taken by the coconut to reach the ground be t s

From the formula

s = u × t + 0·5 × a × t²

where

s is the displacement

u is the initial velocity

t is the time taken

a is the acceleration

In this case taking upward direction as positive and considering the equation in vertical direction

s = -50

u = 0 (∵ initial velocity in vertical direction is 0)

a = - g = - 9·8 m/s²

- 50 = - 0·5 × 9·8 × t²

t² = 100 ÷ 9·8

∴ t ≈ 3·194 s

In this time distance travelled in horizontal direction = 28 × 3·194 = 89·432 m

∴ Horizontal distance travelled by the coconut from the point where it is dropped = 89·432 m

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach