If you understood anything about Physics, which you claimed to understand based on your questions, you would have understood my question right away. It's the most simple Physics question that anybody can ask.
But do me a favor, since I don't claim to be a Physicist like apparently you are claiming to be, due to the nature of your questions posed to me, why don't you simply tell me your answers to those questions? You asked me to answer, so surely you can as well and that's not the least bit unfair of me to ask.
The ball is in your court. I await your excuses.
I talked about well-definedness because you did not state whether or not here is another exterior force acting on the pack of crack you threw away, you did not state either at which angle with respect to the horizonatal direction the object was thrown.
Anyway let's take a simple model because this might be what you are thinking about.
Here are the conditions for a simple ( somewhat well-defined ) model
H1. The mass of the object you threw remains constant in time, we are in the domain of classical mechanics.
H2. The only exterior force acting on your object is gravity.
H3. The initial velocity ( speed and direction ) is V at an angle B with respect to the horizontal line
H4. The path of your object remains in the cartesian plane ( Cartesian came from Descartes a former military and mathematician)
H5. You are a fixed point on Earth( not moving ).
H6. the position in space of the spot from which the object was thrown are ( h, k ) in the cartesian plane of the path of the object, and we start our clock from zero (at initial time) at the very moment when the object is being thrown. The time unit is second.
H7 Everything else is fixed.
Question: How long this pack would remain in the air?
and how do we find this amount of time? The last question is the most interesting one.
Here is the procedure.
Step 1. Use the fundamental principle of dynamics(one of Newton's three laws). Stated verbally this principle goes like this
The sum vector of all exterior forces acting on a object is equal to its mass times its acceleration
Mathematically F= ma
step2.Because the path of the object remains in a cartesian plane the equation is from step 1.
Fx = max
Fy = may
where Fx ands ax are respectively the horizontal components of the sum vector and the acceleration.
FY and ay are respectively the vertical components of the sum vector and acceleration.
step3. based on H1, H2 and H4 and H7 in the hypothesis section we have:
3.1) ax = 0 because gravity acts vertically.( You could desagree but this is a postulate in classical mechanics)
3.2) ay=-g where g = 9.8 m/ square second i.e. g=-9.8m/s/s
" m for meter " and " s for second ".
4.1) Vx= VCosB From step 3.1 and H3 and H7
4.2) Vy =-gt + VSinB from step 3.2 and H3 and H7
Notes Vx = horizontal component of the velocity
and Vy = vertical component of the velocity
CosB = cosine of angle B
SinB = sine of angle B
5.1) X ( the horizontal component of the position of the object )=(VCosB)t + h
Answer: use antiderive and use Step 4.1 and H.6
Y ( the vertical component of the position of the object )= 0.5g( t square)+(VSinB)t + K
Answer: Position is the antiderivative of velocity, and we also use H.6 and H7
where are we heading to ?
Answer: we found the position of the object in the plane. We want to know how long it remains it the air.
step6 Notice that Y is actually the height of the object at time t.
The equation of the height is
Y= 0.5g times( t squared)+(VSinB)t + K .
When the object reaches the ground the Height Y =0
Thus we solve equation 0.5g times( t squared)+(VSinB)t + K =0
It is a quadratic equation,the graph of Y is a parabola.
You can solve by graphing(Using Ti-84) or using the quadratic formula.
For a simplified model set k=0 the answer in this case is
Have a nice day.