A ball is dropped from somewhere above a window that is 2.00 m in height. As it

Question

A ball is dropped from somewhere above a window that is 2.00 m in height. As it falls, it is visible to a person looking through the window for 200 ms as it passes by the 2.00 m height of the window. From what height above the top of the window was the ball dropped?

[When we solved this problem, I don't know why we regard g as positive 9.8, not a negative. ]

The solution is that for 0.200s as it passes by 2m height of the window. s=vt+0.5at2 2 =v(0.2)+0.5*9.8*(0.2)2 v =9.02 m/s height above the top of the window v2 =u2 +2gh.............u=0 h =v2/2g =(9.02)2/2*9.8 =4.15 m

Explanation / Answer

It is not mandatory to take g as positive. As the ball is falling downwards and that direction is taken positive, then g would be positive.

But when upward direction is taken as positive, in that case free falling body would have negative g.

In the given question, v is the velocity at the point ball passes the window, so we take this as final velocity at the window. Now, we have to find the distance above the window. We consider that initially ball is at rest, i.e. u = 0. So, h can be calculated, which is distance from where ball is thrown to the window.

So, You can follow any sign convention, but important point is to remember that you have to follow it throughout.

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