Physics 11 Help?

Science & Mathematics by Anonymous 2018-06-13 20:03:20

Social Science

Physics 11 Help?

3 answers

A stone is dropped from the top of a tall building. It accelerates at a rate of 9.81 m/s^2. How long will the stone take to pass a window that is 2.0 m high, if the top of window is 20.0m below the point from which the stone was dropped?

Anonymous

20*2 = g*t1^2 t1 = √40/9,81 = 2.0193 sec 22*2 = g*t2^2 t2 = √44/9,81 = 2.1178 sec t2-t1 = 2.1178-2.0193 = 0.0986 sec ...a bit less than one tenth of second
Anonymous

s = ½gt² (since initial velocity = 0) t = √(2s/g) Time to fall 20m = √(2*20/9.81) (to reach top of window) Time to fall 22m = √(2*22/9.81) (to reach bottom of window) Time to pass the 2m window = √(2*22/9.81) - √(2*20/9.81) = 0.099 s approx,
Anonymous

as s = 1/2 g t^2 then t = sqrt( 2s/g) The time at the top is sqrt ( 2 * 20 /9.81) and the time at the bottom is sqrt( 2 * 22/9.81) Now could you find the time taken to pass this window from that? Hint :a takeaway problem.