Question

A bomber flies horizontally with a speed of 323 m/s relative to the ground. The altitude of the bomber is 3580 m and the terrain is level. Neglect the effects of air resistance. The acceleration of gravity is 9.8 m/s 2 . How far from the point vertically under the point of release does a bomb hit the ground? Answer in units of m.

Answer 1

Answer:8730.63 m

Step-by-step explanation:

Given

Bomber velocity
`u=323 m/s` relative to ground

Altitude of the bomber
`h=3580 m`

Range of bomb when bomb when it is released is given by

`R=u* t`

where
`u =velocity\ of\ Bomber`

Horizontal velocity of bomber is equal to bomb horizontal velocity

time taken by to cover 3580 m

`h=ut+(at^2)/(2)`

`3580=0+(9.8* t^2)/(2)`

`t^2=(7160)/(9.8)`

`t=27.03 s`

Therefore range of bomb is

`R=323* 27.03=8730.63 m`

Therefore bomber must release the bomb at a horizontal distance of 8730.63 m before the hit point