Question

An airplane is moving at 350 km/hr. If a bomb is

dropped from the airplane at 1.5 km, (a) with what
velocity does the bomb strike the earth? (b) How
long does it take the bomb to fall? (c) What is the
bomb's range?

Answer 1

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

`y=y_(o)+V_(oy)t-(1)/(2)gt^(2)` (1)

`x=V_(ox)t` (2)

`V_(f)=V_(oy)-gt` (3)

Where:

`y=0 m` is the bomb's final height

`y_(o)=1.5 km (1000 m)/(1 km)=1500 m` is the bomb's initial height

`V_(oy)=0 m/s` is the bomb's initial vertical velocity, since the airplane was moving horizontally

`t` is the time

`g=9.8 m/s^(2)` is the acceleration due gravity

`x` is the bomb's range

`V_(ox)=350 (km)/(h) (1000 m)/(1 km) (1 h)/(3600 s)=97.22 m/s` is the bomb's initial horizontal velocity

`V_(f)` is the bomb's final velocity

Knowing this, let's begin with the answers:

b) Time

With the conditions given above, equation (1) is now written as:

`y_(o)=(1)/(2)gt^(2)` (4)

Isolating
`t`:

`t=\sqrt{(2 y_(o))/(g)}` (5)

`t=\sqrt{(2 (1500 m))/(9.8 m/s^(2))}` (6)

`t=17.49 s` (7)

a) Final velocity

Since
`V_(oy)=0 m/s`, equation (3) is written as:

`V_(f)=-gt` (8)

`V_(f)=-(97.22)(17.49 s)` (9)

`V_(f)=-171.402 m/s` (10) The negative sign only indicates the direction is downwards

c) Range

Substituting (7) in (2):

`x=(97.22 m/s)(17.49 s)` (11)

`x=1700.99 m` (12)