# Tag Info

1

In my opinion there is a lot of confusion caused by the fact that the formulas derived for the velocity and height by integrating the acceleration $a=-9.81{m\over s^2}$ often omit the integration constants. These result in an initial velocity and an initial height to be taken into account. Assuming a ball is thrown upwards and leaves the hand at some initial ...

1

The acceleration due to gravity practically does not change near the earth and therefore it is considered constant. If the ball is thrown not high, then it will also accelerate $9.8 \frac{m}{s^2}$ (even astronauts on the ISS are affected by the acceleration of gravity of about $9.2 \frac{m}{s^2}$). How to determine the height to which the ball will rise: ...

1

I'm guessing your confusion is based on a misconception about acceleration and velocity. Acceleration is the rate of change of velocity with time. So as time passes, velocity should change at a certain rate. This rate is the acceleration. So as the ball decelerates, since gravity is acting against it, it's velocity should vary. Hence the velocity of the ball ...

2

My question is when the object reaches maximum height at the point where it is about to turn around downwards, will the acceleration still be 9.8 m/s^2 or will it be greater than this as to help it turn around? It will be smaller, by a small amount. Or a not so small amount if the object goes up several hundred kilometers before falling back to Earth. ...

2

disregarding air-resistance, the acceleration on a projectile (or any free-falling object) is always $\mathrm{g(9.8m/s/s})$. Based on your question, though, I suspect the fact that the object is momentarily “still” vertically at the peak of its trajectory is confusing you. Remember, that magnitude of the acceleration is defined as the  change in the speed ...

1

Acceleration will remain $-9.81 m/s^2$. What changes is the velocity of the ball (think of acceleration as the change in velocity). Whatever initial velocity you give the ball will keep being subtracted by $9.81 m/s^2$ every second, so this means that the velocity will eventually reach $0$, and then start becoming negative. It is at this point the ball ...

3

It will be equal to $-9.8\frac{m}{s^2}$. Acceleration due to gravity is constant, as long as it doesn't stray far away from the surface of the Earth.

2

That movement is ruled by an acceleration, which is the second derivative of position. That means that you only need to specify two things: initial position, and initial velocity, (That includes magnitude and direction). Given the initial position and the velocity, the plane in which the movement occurs is perfectly determined. So you would just choose that ...

0

A method to progress for such a question. Consider a general time $t$ and the position in the space as $r$. At position $r$, draw a tangent plane and a normal plane passing through it. Describe the plane in which the projectile performs a 2D motion (in this case, it will be along the diagonal of the cube). And for projectiles which cannot be described in a ...

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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 ...

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.

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 ...

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There are a two principle factors. The most obvious is physiological. Can the batter do as much work (force times distance) in the action of throwing as he can in swinging the bat. I don't know any clear answer to this, but I strongly suspect that the action of swinging a bat gets more body movement behind it, and therefore more work is done. Especially as a ...

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the batted ball travels faster than the pitched ball because the act of striking it with a bat adds kinetic energy to it. For a perfectly elastic ball, what remains constant is the sum of the kinetic energy of the bat and that of the ball before and after the ball is hit by the bat.

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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 =e×velocity 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 ...

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 ...

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You may want to look up rifling - a bullet is made spin around it axis in order to give it stability. This is actually the main difference between a shotgun and a rifle - the former does not have a helical groove in its barrel and therefore is much less precise (although nowadays the distinction is probably rather blurred).

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The radius of curvature of a curve $y(x)$ is given by the simple formula, $R=|\frac{(1+y^{\prime 2})^{3/2}}{y^{\prime \prime}}|$ where $y^{\prime}$ denotes derivative w.r.t $x$. You know the equation of the trajectory, $$y=x \tan \theta-\frac{g x^2}{2 v_0^2\cos ^2 \theta}$$ Plug in and get $R$ as a function of $x$. Plug in values of $x$ as mentioned in the ...

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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 $t=(\sqrt {2}-1)u/g$

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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 ) $$is correct, but it's only possible for the first and last bullet to land at the same time for a particular velocity. ... 0 The graphs make sense, albeit a bit misleading because of the different scales for red and blue. Friction opposes motion, so if in y=b + m t if b is positive then m must be negative (blue case) and if b is negative then m must be positive (red case). The way I read it, we have$$\begin{array}{l|l} \text{before, t<2.5} & \text{after, t>2.5}...

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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 ...

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Providing “swing” refers to spinning of the ball, the explanation you are looking for is the Magnus effect. A spinning object moving through a fluid (in this case a ball moving through air) will deviate from a parabolic path: a ball spinning clockwise as viewed by the bowler will curve to the right, and the converse for an anticlockwise spin. There is also ...

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