What's Wrong with My Ball Throw Calculations?

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The discussion focuses on calculating the time and velocity of a ball thrown vertically, specifically at heights of 10m and 40m, with a noted time difference of 4 seconds between these points. The initial calculations yielded a negative velocity at 30m, prompting a reevaluation of the approach. It was suggested to consider the ball's motion from the 10m mark to the 40m mark to clarify the initial velocity and its relationship to the final velocity. The equations of motion were highlighted as essential for determining the correct values, emphasizing the importance of the chosen direction in calculations. Ultimately, the issue appears to stem from the assumptions made about the direction of motion and the reference points used in the calculations.
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A ball is thrown verticaly from the ground .Find out the time and velocity when the ball reach 10 and 40m from the ground. The time given between 10m and 40m is 4s.
I tried to solve this question by asumming that t30 = t10 + 4s and I found that the value of v30 is negative. So I try another way to solve it. This time I do not assume t30 = t10 + 4s. I find out all the values besides t first and finally I got the value of v30 is positive . However, t30 is not equal to t10 + 4s .

What`s wrong actually??
 
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Nothing may be wrong. If you chose the positive direction downwards, then v will be negative. What equation did you use?
 
Try assuming that the ball was thrown from the 10m mark up to the 40m mark instead of from ground. You know the distance between (40-10), the time between, and the acceleration. All you need is the initial velocity. You should have an equation for that. Once you find the initial velocity (which, from what you just solved is actually the velocity at 10m), this velocity becomes the final velocity. You also have the distance and acceleration. You need to then find the initial velocity (this time referring to the ground). Good luck.
 
To find the velocity at d=10m use

change in displacement = v(initial)t + .5at^2


From there, you can find the velocity at d=40 from

v(final)^2 = v(initial)^2 + 2a(change in displacement)
 
I think there must be something wrong...since i get positive value of v10 but negative value of v30.
 
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