[When we solved this problem, I don\'t know why we regard g as positive 9.8, not

Question

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

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

while passing the window

velocity at the top of window = v1y

displacement y = -2 m

time t = 200 ms = 0.2 s

acceleration ay = -g = -9.8 m/s^2

y = v1y*t + (1/2)*ay*t^2

-2 = v1y*0.02 - (1/2)*9.8*0.2^2

v1y = -9.02 m/s

the ball is dropped from a point h above the top of window

displacement of ball after reaching the top of window y = -h

initial velocity at point h , voy = 0

v1y^2 - voy^2 = 2*ay*y

(-9.02)^2 - 0^2 = -2*9.8*(-h)

h = 4.15 m

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