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The equation of motion for the ball from the time it bounces till the time it hits the ground again is $$ y = v_0t - \frac{1}{2}at^2 $$ where ground level is $y=0$, and $v_0$ is the velocity going up after adjusting for the coefficient of restitution, and $t$ is the time since the bounce. This equation will take the ball through its peak and back to the ...


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Your formula's wrong. You've got $v=\frac12 at^2$, whereas that's the formula for $y$=height. Velocity's actually $v=at$ (with $a=9.8\mbox{m/sec}^2$).


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If the lift is going up or down maintaining constant velocity or if it is not moving at all, in all the three cases if you drop the ball it will take the same time to hit the floor. It is freefall in all three cases because you are just dead-dropping the ball and not giving any initial velocity or accelaration. (Inertial Frame) What you are referring to ...


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Only if the situation is symmetric and the masses are aligned like an equilateral triangle all masses will meet each other at the same time: If the initial distances are not the same the point masses will miss each other and start orbiting around each other. For example the initial configuration of an isosceles triangle: Here the two smaller masses ...


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In a vacuum the rod will keep its orientation because of the equivalence principle, but with air resistance the heavier part will tilt toward the ground because its mass and therefore force is higher. The force depends not only on the mass of the planet but also on the mass of the falling object, and since air resistance is also a force that acts on the ...


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The basic equations, assuming no air drag($^*$) are as follows. At $t=0$ we drop the object from height $H$, we assume its initial speed is zero ($v_0=0$). Only gravity ($mg$) is acting on it. If the object free falls to height $h_1$, energy conservation then shows us that its speed has now become: $$v_1=\sqrt{2g(H-h_1)}$$ At height $h_1$ a braking ...


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If you take down as positive then displacement $s = +30$ m, initial velocity $v_i = +8$ ms$^{-1}$ and acceleration $a = +10$ ms$^{-2}$. Using the constant acceleration kinematic equation $s = v_i t + \frac 1 2 a t^2$ where $t$ is the time gives $$(+30) = (+8)t+\frac 1 2 (+10)t^2$$ If you take up as positive then displacement $s = -30$ m, initial velocity ...


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The only condition for free fall as you said is that the motion of the body should be only under the influence of gravity alone. There should not be any effect of other forces like air resistance, viscous drag etc. The condition depends on the property of the material under free fall. For example, if the body has a certain mass as well as charged, it causes ...


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In the quote cited you could imagine that point particles move in a straight line at a steady velocity and don't rotate when far from massive objects. So to a small accelerating frame, they look like a non rotating point particle moving at a constant velocity would look to an accelerating frame. But maybe point particles near a massive object have similar ...



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