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Could someone please explain the trajectory of the ball that is bouncing in this picture...

The vertical component of the velocity of a bouncing ball is shown in the graph below. The positive Y direction is vertically up. The ball deforms slightly when it is in contact with the ground.

Ball track

I'm not sure what the ball is doing and when, what happens at 1s?

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6 Answers 6

I am late so as said by other's the graph just plot's the velocity of the ball with respect to time.

So what the graph say's. as said it is the plot of velocity vs time so as you should know velocity can be either negative, positive and 0 but speed can't because velocity is a vector quantity and it have magnitude as well as the direction unlike speed which only have the magnitude that's why we have a negative side in the graph also

$Analyzing$ $the$ $graph:-$

You didn't mentioned but i am assuming and also the graph say's that the ball is released from certain height toward's the ground in a free fall.

Now let us assume the velocity toward's the ground to be negative and toward's up to be positive

So at $t=0$ we release the ball and it start's falling and gaining speed and reaches to a speed of $9 m/s$ in $1 sec$ which is due to acceleration $g$. Now it is shown $-9m/s$ in the graph because we assumed and your textbook too that the velocity in downward direction to be negative

Now at $t=1s$ the ball strike's the ground and bounces backward gaining the velocity of $9m/s$ instantly as you also might have experienced while playing with a ball in the ground that when you throw it toward's the ground it bounce's back instantly toward's you. Now you might see a little time gap of few millisecond in the change of velocity from $-9m/s$ to $9m/s$ is because that that time was wasted in the deformation of the ball when it strikes the ground

Now the ball start's rising upward's at $t= 1.5s$ till it reaches to the top and stop's due to the deceleration $-g$ acting on it and reaches to $0$ velocity at $t=2.5s$ and again start's falling from the top and gain speed and same cycle goes again

Hope it helped you

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Remember $$v=\frac{\text{d}x}{\text{d}t}$$ Then $$x=\int_0^t v(t) \text{d}t$$ Simple integration gives you the following plot of $x(t)$:

enter image description here

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Initially, the ball is at rest. Therefore, the initial velocity is zero as mentioned in the graph. The negative gradient states that the ball has a downward acceleration. The moment, it hits the ground, the velocity reaches to its maximum. Then, it starts accelerating in the upward direction. The graph indicates a larger upward velocity due to the deformation and reformation of the ball.

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This is a graph of velocity versus time, so the slope of the graph at any point gives the change in velocity over time, or dv/dt which is acceleration. Acceleration occurs due to the application of force to a mass (f=ma, or a=f/m).

Any change in the slope of the graph must occur due to a change in (dv/dt), which is a change in acceleration, and hence a change in the applied force. The only forces at work here are gravity and the "spring" like force caused by deformation of the ball. The changes at 1s and 1.25s are due to the addition and removal of the "spring" force from the deforming ball as it strikes and then leaves the ground. Gravity applies throughout the graph. Both these forces are "constant" where applied, giving uniform acceleration and hence a straight line dv/dt graph with some slope.

Velocity-time graphs can be slightly confusing in that a change in direction of motion, something we would see as a major event, is indicated only where the line crosses zero. If this change of direction is due to constant acceleration, like a ball at the top of its arc or at the transition of a spring-like deformation from compression to expansion, this occurs in the middle of a line of constant slope (2.25s and 1.125s above), and seems unremarkable. Crossing zero is always a point of potential interest in a graph.

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You need to remember that this isn't a plot of position vs. time, it's velocity vs. time. Now that that's emphasised, let's analyse the plot.

Clearly, the ball starts out at zero velocity and the velocity is increasing in the negative direction linearly with time. But we also know that the acceleration on a body is given by $$ a = \frac{dv}{dt}$$ i.e., the rate of change of velocity. I don't know familiar with calculus you are, but for a straight line the slope ($dv/dt$) is constant. Therefore we can infer that the acceleration is also constant.

Do we know a physical situation where this happens? Sure... a ball released from a height above ground will follow exactly that motion, with the constant acceleration being provided by gravity.

For what happens at $t = 1$ $s$ - The velocity of the ball goes from $-9$ $m/s$ to $9$ $m/s$ really quickly. This just means that the direction of the velocity changed, without changing the magnitude. This happens when a ball bounces off the ground, neglecting friction and other losses. So that's your system - A ball bouncing off the ground.


P.S - There's another way to see that this is a ball falling - Acceleration is the rate of change of velocity. And we know that the acceleration due to Earth's gravity is ~ $9.8$ $m/s^2$ so in $1$ $s$ the velocity will be ~ $9.8$ $m/s$, which is roughly what the graph is showing here.

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At $t_0$=1s, the ball bounces in a perfectly elastic manner and there is no loss of momentum. Therefore, $v(t=t_0+dt) = - v(t=t_0)$. If the recording had a finer resolution in time, this would show up as a vertical jump but here you have a milder slope since it is recording only every 0.1 s. Between the bounces, gravity takes over and the ball accelerates downwards. If you compute the slope of the decreasing part of the curve, you should find a well-known constant.

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