# Velocity time graph for a ball thrown vertically upwards and falling back down to be caught again by a person

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.

• 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. Dec 15, 2015 at 17:26
• 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. Dec 15, 2015 at 18:54