12

This is a great example about limitations of mathematical models applied to physics. For solving the problem, you use the very well know, quadratic equations for the motion of the ball. This means that the resulting motion is a parabola, and then math says that there are 2 different solutions for the problem. Now, after doing the math, you have to put that ...


3

The t = -1 second root to the equation represents the case where the ball was thrown from zero initial height instead of from the top of a building. It's the second intersection point of the whole parabola with the ground. You exclude it because it has the wrong initial conditions for your particular problem.


2

I can't really speak about physiological differences that might influence the exerted forces. But the main differences as far as the ball is concerned are: Batting is an elastic impact, while throwing can be thought of as an inelastic "impact". The impact speed is larger when batting When throwing a ball, you get it to exactly the same velocity ...


2

It cannot be done with a mathematical model, because swing bowling (quite distinct from spin, btw) depends on the layer of turbulence adhering to the edge of the ball, and we cannot accurately give a mathematical model of turbulence. However, the basic physical principles can be explained. First, consider golf balls. Golf balls are dimpled because it was ...


1

That's why physics is not mathematics, because you have to study boundary conditions and validate them ! In general body movement in $x$ coordinate axis is described as : $$ x = x_o + v_o t + \frac{at^2}{2} $$ Noticing that body total distance will be zero when it will touch the ground and that gravity acts in opposite to body initial speed gives the ...


1

A greatly simplified ans comes from the fact that coefficient of restitution(e) is smame for both cases ( since same bodies are colliding). Velocity of separation =eƗvelocity of approach When he throws the ball himself it is usually at very low speeds(ie a low velocity of approach) compared to a baller who put his all in throwing the ball. Velocity ...


1

I dont know why you took the derivative (plus I think you might have a typo in wolfram alpha), but after you get $t + \frac{2u \sin \theta}{g} = \frac{2u \sin \frac{\pi}{4}}{g}$, you can get $\theta (t)$ straight from there as: $$\theta (t)=\sin^{-1}(\sqrt{2}/2-gt/2u)$$ this is a continuous function and passes through $\pi /6$, when $t=(\sqrt {2}-1)u/g$


1

I got this solution ? with: $$y=v_0\,\sin(\varphi)\,t-\frac{g\,t^2}{2}\tag 1$$ $\Rightarrow$ $$y=0\quad,t_0(\varphi=\varphi_0)=2\,{\frac {v_{{0}}\sin \left( \varphi _{{0}} \right) }{g}}$$ for next bullet: $t\mapsto t_0-dt$ and $\varphi\mapsto \varphi(t)$ in equation (1) $$y_t=v_{{0}}\sin \left( \varphi \left( t \right) \right) \left( 2\,{ \frac {v_{{0}}\...


1

I was able to find the issue and apparently (according to meta) its OK to post an answer to my own question. The working in the question leading to the solution of $$ \alpha(t) = \sin^{-1} \left (\frac{1}{2} + \frac{g}{2u}(1-t) \right ) $$ is correct, but it's only possible for the first and last bullet to land at the same time for a particular velocity. ...


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