# Kinematics: 2 body problem (velocity and acceleration) [closed]

I have a kinematics problem, and it goes like this.

A superhero, Ironman is flying upwards at a constant velocity of v = 20m/s. At a certain time a spherical tool he was carrying falls from his hand. (note: imagine the spherical tool is just a metallic ball) For the next 2 seconds the ball falls under the influence of gravity. Then due to air resistance, reaches terminal velocity at 2seconds.

I wish to calculate the constant velocity it reaches after 2 seconds and the distance it fell in the 2 seconds.

If I try to calculate its displacement/distance, does this ball have an initial velocity after it gets let go.

d = v1*t + (1/2)gt^2 ( where g = 9.8m/s^2)

Will v1=0, or will it be equal to v = +20m/s or -20m/s.

Hope someone can provide some guidance!

## closed as off-topic by ZeroTheHero, Chris♦, ja72, Emilio Pisanty, sammy gerbilFeb 23 '18 at 15:13

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• Please clarify what do you mean by constant velocity – QuIcKmAtHs Feb 22 '18 at 21:48
• Hi, constant velocity, that is standard terminology in Physics. That means that as one travels, your magnitude and direction of the velocity does not change. I hope this makes it clear. – Palu Feb 22 '18 at 23:44
• -1 Not clear. The problem states that there is air resistance, but you have not included this in your calculations. – sammy gerbil Feb 23 '18 at 15:16
• Hi, the terminology i am using is standard physics as far as i am aware of. This is how its stated in the books in general. There is air resistance such that this results in the object achieving Terminal Velocity, and this term means we reach Constant Velocity. SO there is no Real Friction to take into consideration in terms of calculations. I just can't believe that my terminology which i feel is very standard is causing so many issues here. This is terminology that is standard in North American English books. So – Palu Feb 23 '18 at 19:14
• In projectile motion questions, air resistance is taken to be zero. If you do not believe as such, please sketch for me how you would think the motion of the object would look like – QuIcKmAtHs Feb 24 '18 at 1:49

Since iron man is travelling up and drops the object (presumably straight down) the problem is entirely in one component only. The pitfall in solving this problem can come from assigning appropriate sign conventions.

So for example, suppose you orient positive to be in the direction of Iron man's flight, then the initial velocity will be +20 and the acceleration due to gravity must be -9.81.

The rest, as you have outlined, is just kinematics.

• Hi Tarik, your interpretation I believe is correct, so you also agree that as soon as the ball is dropped, at that instant, it does have initial velocity then of +20 given that up is positive convention and down is negative convention. So the equation for displacement then is: d=v1*t + (1/2)*g*t^2. SO with the values: d = 20t +(1/2)(-9.8)*t^2 or rewritten as: d = 20t - (4.9)*t^2. – Palu Feb 22 '18 at 23:37
• Correct, with time being t=2 as well. – user185532 Feb 23 '18 at 0:15
• OK, thank you Tarik. I will choose your answer as the correct one. Thanks for clarifying. – Palu Feb 23 '18 at 0:21

When Iron Man is moving at $20ms^{-1}$, every single object on him would move at $20ms^{-1}$. Hence, the immediate moment when the ball leaves his hand, the ball would also be travelling at $20ms^{-1}$.

However, you approach is wrong here. This question is about projectile motion, and you should involve the x and y components of the velocities and acceleration.

• Hi, this is not a projectile motion problem, the ball just drops straight down. – Palu Feb 22 '18 at 23:33
• How is that possible? The ball possess momentum! – QuIcKmAtHs Feb 23 '18 at 4:38
• when I say its not projectile motion, it due to the fact that it is only in 1-dimension up and down. Most people would regard projectile motion when the projectile is traveling in a parabolic arch, that is in 2-dimensions. – Palu Feb 23 '18 at 14:26
• When the object leaves iron man, it will fall down at a certain velocity, but also move forward at 20ms^-1 – QuIcKmAtHs Feb 24 '18 at 1:48
• Yes, thanks for stating this idea QuIcKmAtHs. Appreciate it greatly. – Palu Feb 24 '18 at 5:30