0
$\begingroup$

I am currently studying AS level physics and having some difficulty understanding the shape of a velocity time graph that represents the motion of a ball being thrown upwards and falling back down to be caught again (https://www.google.co.th/search?q=velocity+time+graph+of+an+object+being+thrown+upwards&client=ubuntu&hs=CJZ&channel=fs&source=lnms&tbm=isch&sa=X&ved=0ahUKEwi_wLCZm97JAhVTG44KHWddBgcQ_AUIBygB&biw=1920&bih=943#imgrc=Qa7PekAU2A8XzM%3A) here is a link to the image. The person throwing the ball would need to give it a force which would causes the ball to accelerates but doesn't this takes time? if say we workout the value of acceleration using the equation F=MA to be 20m/s/s isn't it true that it would take the ball 1 second to reach a velocity of 20m/s so why when you look at the graph (from the link I provided) it says that at time = 0 the ball has a velocity of 20m/s when clearly at time = 0 the ball should be at rest, and because this information is not on the graph then what happen to the displacement (area under the graph)? during that time, does this mean it has been ignore?

Thank you.

$\endgroup$
1
$\begingroup$

Yes, the interaction with the hand seems to have been ignored, and the graphs are only considering what is happening immediately after being thrown up to immediately before being caught. It is typical to ignore that because modeling the flight under the influence of just gravity is much easier for introductory level physics. You'd have to add quite a bit more information to discuss what happens when the ball is thrown or caught.

$\endgroup$
  • $\begingroup$ When the ball's velocity increases from 0 to 20m/s after 1 second (because acceleration is 20m/s/s) didn't the ball already left the hand? I am confuse to what you mean by "immediately after being thrown" because surely it should gain distance during that 1 second before reaching a velocity of 20m/s $\endgroup$ – CuriousGuy Dec 15 '15 at 17:22
  • $\begingroup$ It would if it were considering those things. But that graph is considering $t=0$ to be the time at the release of the ball. At that point, it sets the initial position to 0, and the velocity has already reached $v=20 \text{ m/s}$. Once the ball leaves the hand, no more force is applied by the hand, and the ball decelerates at $g$. That graph simply doesn't consider anything that happens during the interaction with the hand. $\endgroup$ – tmwilson26 Dec 15 '15 at 17:26
  • $\begingroup$ I have gain a better understanding of the graph now just a few questions left. Because the graph ignore the change in velocity of the ball from 0 to 20m/s during the first second is it safe to say that it ignores the distance gain by the ball during this first second as well? Secondly, t = 0 is the time at which the ball left the person's hand right? and what do you mean by sets the initial position to 0? $\endgroup$ – CuriousGuy Dec 15 '15 at 18:50
  • $\begingroup$ Thats right, it is ignoring the distance change during that time period as well. You are correct on the second point as well, $t=0$ is the time of release. Position 0 on the graph is the position of the ball right when it is released, not the position right at the start of the hand motion. It just means that the graph of position starts at 0 when the velocity is 20 m/s. The main point of those graphs is to show that the velocity is linear under the influence of gravity, and that the position is quadratic. $\endgroup$ – tmwilson26 Dec 15 '15 at 18:54
  • $\begingroup$ So if 20m/s is the velocity of the ball when it leaves the person's hand then what is the velocity at which it will hit the person's hand once it falls back down? and does newton's third law have anything to do when the ball hit the person hand? $\endgroup$ – CuriousGuy Dec 15 '15 at 19:11
0
$\begingroup$

I think the question is more elementary than you make it out to be, the main purpose being to test understanding of the fact that the velocity/time dependence is linear, so that the graph is a line (as opposed to the position/time graph, a parabola).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.