I don"t acquire it, why is the graph constantly linear? Shouldn"t that slope be zero when the sphere is in ~ the top and stops there for a moment?The means it is "explained" is that it"s in ~ the optimal for instantaneous moment. Okay, but, if let"s say, round was going upward for a 5 sec, then stopped for 1 2nd and walk downwards. Wouldn"t girlfriend agree v me, that if we were to record the ball"s motion, and also compared that velocity in between 3 and also 4 seconds, the distinction would be much higher than if we contrasted velocity in between 5 and 6 secs (when the ball"s velocity isn"t changing, it merely hangs in the air, thus there would be no difference in velocity between 5s and 6s)?? If so, then it have to be reflect on the graph, considering that the scale is fixed.

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When the round is in ~ the top and also stops for a moment, the velocity the the round is zero. This is represented as the suggest where the velocity graph crosses the time axis.The slope of the velocity graph, however, describes how the velocity is changing, and has naught to carry out with if the ball is stationary or moving. Zero velocity means the velocity graph is in ~ zero; the does not median the graph"s steep is zero.The steep of the position graph in ~ this moment would certainly be zero, yet not that of the velocity graph.
Not for a ball thrown up. In a different scenario you might have zero steep somewhere. For example, the velocity function of an elevator. Also, if you incorporate air resistance, the velocity graph for the round will come to be non-linear, yet you still won"t acquire zero slope at the turnaround point, since the drag is zero there. Just if the sphere would method terminal velocity as soon as coming down, the graph would technique the horizontal.
The ball doesn"t precise stop. It appears to be stationary because that a portion of a second when the direction that velocity changes.The velocity axis (y-axis) mirrors whether the sphere in going increase or down. (positive : round going up, an unfavorable : round going down)The minute when the direction the velocity alters is the one wherein the magnitude of velocity is zero (where it cut the x-axis)
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If you store making the time that it"s stationary shorter and shorter you end up with your original drawing.
Yes. Agreed. Yet dramadeaur was a particular amount that time for which the velocity would remain 0. I beg your pardon would mean the graph would be horizontal at v = 0 because that a tiny time. However that isn"t the case. Just wanted come say that.
View attachment 78170If the sphere stopped because that an even much shorter time it would certainly look choose the one below.View attachment 78171If you save making the moment that it"s stationary shorter and shorter you finish up v your original drawing.
If the sphere was walking upwards for 6 seconds (at 6th second that reaches the top) does it mean that the change in velocity in between 4-5 secs would be the exact same as the adjust in velocity in between 6-7 seconds?
Does it median that the adjust in velocity in between 4-5 seconds would it is in the same as the adjust in velocity between 6-7 seconds?
Yes, as currently sated multiple times here, acceleration due to gravity is consistent (on such little scales).
I don"t really recognize the bottom component of the graph. Together what you guys told here, i recognize the upper part of the graph. The top graph of the graph indicates that the ball is thrown upwards and reached the best height.But then how about the lower part?? is the sphere bouncing upwards again or the ball is just collapsed??
I don"t really recognize the bottom part of the graph. As what you guys told here, i recognize the upper component of the graph. The upper graph of the graph indicates that the sphere is thrown upwards and reached the preferably height.But then how around the lower part?? is the ball bouncing upwards again or the round is just collapsed??
Note that it is a graph the VELOCITY not SPEED. The graph assumes the up is positive. In ~ the begin the ball is going up through some early stage velocity. That velocity reduces until it will zero at the top. The ball then falls. Top top the way down the velocity is as such negative.At some allude it hits the ground v the very same SPEED that it was launched with however the VELOCITY will certainly be negative.
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Likessophiecentaur
Note that it is a graph the VELOCITY not SPEED. The graph assumes that up is positive. At the start the round is going up with some initial velocity. The velocity reduces till it get zero at the top. The sphere then falls. ~ above the means down the velocity is therefore negative.At some suggest it hits the ground through the very same SPEED the it was released with however the VELOCITY will be negative.
Interpreting graphs is not an innate skill. A diminish line suggests downward movement to someone who does"t instantly read the brand on axes. I defy anyone, even someone with years of suffer of using and also interpreting graphs, to claim that, even now, they constantly get that right an initial time. Using the Area under a graph to present a street travelled, is another real hurdle because that the uninitiated.Give students the job of drawing graphs of position, velocity and acceleration for a bouncing superball. You space guaranteed to obtain some bizarre results.

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Note that it is a graph the VELOCITY not SPEED. The graph assumes that up is positive. At the start the round is going up with some initial velocity. That velocity reduces until it reaches zero in ~ the top. The ball then falls. Top top the means down the velocity is therefore negative.At some suggest it hits the ground v the very same SPEED the it was released with yet the VELOCITY will be negative.