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You need to research the mechanics of drag and the Drag Co-efficient (start with this wiki page). A simple model (where drag is a Ram Pressure) holds the drag to be proportional to the square of the speed. This can be justified on simple momentum conservation grounds, in the case of pure ram pressure. The drag co-efficient is an empirically-found "fudge ...


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Your want to add a drag term to your force equation. It will be more complicated and involve the geometry of the objects in question. Note also your supposition will not always hold, an adult in a parachute will not fall faster than a child without a parachute. {disclaimer: Not tested empirically!!!!} You will have drag force term like: $$\vec{F}_{drag}\...


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If I read the description correctly, your experiment is set up to impart an initial velocity to the coin before your timing starts - as you drop it from slightly above the edge of the table, but don't start timing until you hit the edge (hear the impact). At the moment of impact, the center of mass of the coin will slow down a little bit - but it will still ...


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You must look at all forms of energy. Just before the explosion, the projectile has gravitational potential energy GPE, kinetic energy KE, and also chemical potential energy CPE stored in the dynamite. Just after the explosion the 3 fragments all have the same GPE as before. The CPE has disappeared in the explosion. As Jim says, we must assume that it ...


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You don't actually have to use harmonic motion,you can solve this question by using equations of motion. Given that the collision are elastic (coefficient of restitution is 1),i.e,when the ball collides with the building its velocity is reversed. The point to note here is only the horizontal component is affected by the collision. Along the ...


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You already have a relation between $\theta$ and $\alpha$. If this is correct, all you need to do is to rearrange it. If it is not possible to rearrange it to get the given relation, then either your equation is wrong, or the given equation is wrong, or both. If the particle grazes the upper plane at maximum height, where the velocity vector is horizontal,...


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Because you can separate the vector space (assumption) of R x R into two separate vector spaces, each spanning the real line R with independent basis. When you are in the vector space R then you can't reach all elements in R x R. You can do that if you make a superposition of each of these R spaces into R x R and then you can reach all elements in this space ...


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The above isn't drawn correctly. And it is far too complicated in my opinion. I think I was able to solve this using high school physics: This bugged me a long time. I am surprised no one ever posted an answer on the internet. The length of his rope was 60.8 meters, assuming a drop above the target building of about 12 meters. The difference in height ...



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