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If we model the motion of a bouncing ball on a velocity time graph, neglecting air resistance, we get a line with gradient -9.8 as the ball falls , and then a steep positive gradient line when the ball hits the floor and then the same -9.8 gradient line again as the ball goes up. What I don't understand is if we take the downwards direction as positive, then shouldn't the ball have a gradient of 9.8 as it falls and then a gradient of -9.8 as it goes up ? The same question for acceleration time graphs as well. Moreover why is there a sudden steep gradient at the impact with the floor. I hope someone can talk me through the velocity-time and acceleration-time graphs for a bouncing ball neglecting air resistance. My calculus skills are limited, but I do understand the concepts of differentiation and integration and how velocity is the derivative of displacement and so on.

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    $\begingroup$ Might help if you tried to better describe the conceptual problem that you're having by providing a velocity-vs-time graph of what you think the motion should look like. $\endgroup$
    – user93237
    Sep 21 '17 at 20:08
  • $\begingroup$ Your graph is incorrect as the acceleration of the ball due to the gravitational attractive of the Earth does not suddenly reverse in direction after the bounce. What you need to draw is a velocity against time graph which is of constant gradient $\pm g$. The sign of the gradient will depend on the direction (up or down) which you chosen to be positive. $\endgroup$
    – Farcher
    Sep 21 '17 at 20:47
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Your diagram is incorrect. The objects always experiences negative acceleration, except for the brief moment when it bounces. The negative acceleration always reduces the speed (taking it from positive to negative), but during the impact the speed abruptly switches from negative to positive. The position is what you expect, with the object bouncing up and down in a parabola vs. time.

I drew a crude acceleration (red), velocity (blue) and position (orange) chart below for illustration.

pva

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  • $\begingroup$ @RobertDiGiovanni - as long as it is in contact there is a contact force, and the contact force accelerates the ball upwards regardless of the bounce stage. Well kind of, because if the contact force is less than the weight then it still accelerates downwards. Anyway, as I said in my post, these are crude curves to drive a point. $\endgroup$ Dec 12 '20 at 20:54
  • $\begingroup$ Interesting point. Yes, one is accelerating in the opposite direction when slowing down. I should have said move in the opposite direction after maximum compression. $\endgroup$ Dec 13 '20 at 2:45
  • $\begingroup$ But it won't "bounce" unless some type of "spring" compresses from the collision. A tennis ball bounces of a hard floor, but not off of sand. $\endgroup$ Dec 13 '20 at 2:52
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Your acceleration diagram is not correct. The acceleration due to gravity is always in the positive direction regardless of the value of the velocity. First, the acceleration will be +9.8 then it jumps up to infinity (for an instantaneous change of momentum for the bounce) or a very high negative value for a slower change of momentum. Then it jumps back to +9.8 because the object continues to accelerate in the positive (down) direction after the bounce.

The slope of the v-t graph is acceleration. The gravitational acceleration is always in the same direction, the positive direction as you have it set up. The gravitational force doesn't care what the velocity is, it will always change the velocity in the positive direction creating a positive acceleration and a positie slope in the v-t graph. You can start a ball with any velocity at all, under gravitational force the velocity will change to become more positive giving it a positive slope. So velocity starts from 'vi' (the velocity the ball is dropped at), and begins to increase in the positive direction (down). So as time progresses forward, the ball picks up speed in the positive direction, the direction the velocity is in. So for the gradient we have +v/+t for +9.8m/s. Now when the ball hits the ground with final velocity 'vf' it rebounds (lets assume a perfectly elastic collision). That rebound from vf to -vf forms a straight vertical line from t='time it hits the ground' from vf all the way up to -vf (if the impulse is infinite and the change in momentum is instantaneous). If the change in momentum is not instantaneous, there will be a line of steep negative slope from vf to -vf where the slope is 'the change in momentum'/'time it took'*'mass.'Then the velocity begins to increase from -vf as the ball bounces up. I mean, the absolute value of the velocity decreases, but the velocity technically increases as it moves away from -vf becoming less negative. The important part is that the acceleration is now fighting the velocity and it is in the opposite direction of the velocity but the velocity is becoming more positive now, so it still has a positive slope of +9.8.

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x-t,v-t,a-t

A ball is dropped and its displacement vs time graph is as shown in figure (displacement $x$ is from ground and all quantities are +ve upwards). Displacement, $x$ from the ground is positive. All quantities are positive upwards.

As the ball is dropped from some height, its velocity in downward direction is increasing till hitting the ground. During collision velocity of the ball changes the direction. This is shown above.

Acceleration on the ball is constant and always in downward direction so it would be negative. $a=-g$ for all time.

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