Physic: bomb trajectory calculation

Fine. Now this may be a bit confusing yet gravity as we know it will drop things at 32x32 feet per seconds squared until terminal velocity is reached. An interesting sort of fact. Imagine if you will, say standing on a ladder. You have a bullet or something like that and then you drop it. Well gravity being what she is whatever that thing is is going to hit the ground in X amount of time. Basically this is true of most objects Now this gets even more fun if you were to take in other factors like wind, aerodynamics, earth rotation and things like that. it really can become sort of complex.

Well,
to push something aerodynamic shaped or not at mach 1.5 ( which is 1.5 the speed of sound ) through air takes quite some energy. My guess that the bomb will slow down horizontally to some 50-100 km/h within the first few seconds also the acceleration will rather soon will be compensated by the drag to something below 300 km/h.
See drag goes with v ^ 2 !
And aerodynamics above mach 1 is quite a different thing.
nice comparing animation there ->

At that speed you cant open a door to drop something, because it would rip the opening until the aircraft “disintegrates”.
I think systems used with that kind of planes are active driven missiles which can go way faster than mach 1.5 and may reach anything the pilot can spot.

Just for the record, thats faulty - but irrelevant for the calculation.
Mach depends on the speed of sound, which depends on the temperature of the medium its moving in.

for 20°C it would be 343ms = 1235km -> 1593/1235 = Mach 1.28
at 15km height it should be around -50 to -70°C and the speed of sound should be around 1078km/h to 1024 km/h resulting in Mach 1.48 to 1.54.

:smiley:
wrong again.

1 km / 1 h = 1 km/h
1 km = 1000m
1 h = 60 min = 3600s
1 km/h = 1 * (1000)m/(3600)s
1000/3600 = 0.277777 ~ 0.278
1 km/h = 0.278 m/s

you might also need to take into account that the earth is round, and at such an altitude and speed the bomb might enter a slight orbit and thus travel further than otherwise expected. (not sure how much further that would be)

I don’t think this is correct due to the fact that there is no force acting against the motion of the bullet travelling around the earth’s centre that would not be acting against the air, the surface of the earth or anything else you can think of. Therefore, relative to us the bullet would just fire straight up, and the only force acting against it would be the force of gravity bringing it down.

I could be wrong here, but I’m willing to bet that this is correct.

this is not 6th grade math.

the simple formula is:

height in meters = (9.81 * time in seconds^2)/2

so…

15000 = (9.8 * time^2)/2
215000 = 9.8time^2
(215000)/9.8=time^2
sqrt[(2
15000)/9.8]=time
55.328…=time

velocity is 1593 km/h = 1,593,000 meters per hour = 26550 meters per minute = 442.5 meters per second

time * velocity = distance
55.328seconds * 442.5meters per second = 24482 meters

Great. In theory only!!!

the velocity at any given time is:

velocity = gravity * time
v = 9.8*55.328 = ~542 meters/second

the speed of sound is 343 meters per second
do you really think that the bomb is going faster than the speed of sound when it hits??? No. It reaches terminal velocity before that.

So you have to calculate terminal velocity before you can do the math.

Show me a 6th grader who can do this. Ha.