Question

Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 688 m/s (mach 2) . They release their bomb and head for home. How long will it be before the bomb hits the ground?

Answer 1

The bomb will remain in air for 17.5 s before hitting the ground.

Step-by-step explanation:

Given:

Initial vertical height is,
`y_0=1500\ m`

Initial horizontal velocity is,
`u_x=688\ m/s`

Initial vertical velocity is,
`u_y=0(\textrm{Horizontal velocity only initially)}`

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is
`S=y-y_0=0-1500=-1500\ m`

Acceleration in the vertical direction is due to gravity,
`g=-9.8\ m/s^2`

Therefore, the displacement of the bomb is given as:

`S=u_yt+(1)/(2)gt^2\\-1500=0-(1)/(2)(9.8)(t^2)\\1500=4.9t^2\\t^2=(1500)/(4.9)\\t=\sqrt{(1500)/(4.9)}=17.5\ s`

So, the bomb will remain in air for 17.5 s before hitting the ground.