Projectile Motion: Solving for Meeting Point of Projectile and Falling Object

AI Thread Summary
A mathematical proof is sought to demonstrate that a projectile and a falling object will always meet, regardless of their initial speeds or ranges. The discussion emphasizes that both objects experience the same gravitational force, leading to a meeting point after a certain time. The key to the solution lies in understanding the relationship between horizontal and vertical motion, where the horizontal component of the projectile's speed must be sufficient to cover the distance before the falling object reaches the ground. The time of fall can be calculated using the formula Height = 1/2*g*t², and as long as the projectile's horizontal speed exceeds the distance divided by time, they will collide. This concept can be illustrated graphically to show the equivalence of vertical components and the timing of their intersection.
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Homework Statement



Prove mathematically that regardless of the range and initial speed, a projectile and a falling object will always meet.

Homework Equations



Vf^2 = Vi^2 + 2ad, d = vi t + 1/2 at^2, d = vt.. etc basic constant velocity and acceleration formulas.

The Attempt at a Solution



this question followed one with specific information given (which I was able to easily solve) and I know how to explain the answer in words; a projectile is under the same force of gravity as th e falling object, so after a certain amount of time the horizontal component of speed will meet up with the falling object, obviously if there is less distance and/or a higher horizontal speed this will occur quicker, but how do I show this conclusion using mathematical formulas?
 
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If you are in the frame of reference of the falling object, and it is aimed at you and is fired at the moment you drop, then it will strike you straight away, because you and the projectile are moving together. From your point of view it will be traveling straight at you the whole way.
 
LowlyPion said:
If you are in the frame of reference of the falling object, and it is aimed at you and is fired at the moment you drop, then it will strike you straight away, because you and the projectile are moving together. From your point of view it will be traveling straight at you the whole way.

I understand the concept in words but I can't come up with a mathematical equation to prove it. Do I just draw a graph to show that the vertical components are the same and that the amount of times it takes to reach the object is just the horizontal component calculated via d = vt?
 
You can note that in the accelerating frame of reference of the object/projectile world there is no net acceleration on either relative to the other.

You know the time to fall is given by Height = 1/2*g*t2
And the distance is D

So long as the projectile has a V greater than the D/t then it should strike before the ground intervenes to interrupt things.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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