physic
2007-09-27 3:43 am
an object is thrown with a speed of 50m/s straight up from the top of a building which is 100m high. determine the disance, d, of the object above ground level 11 seconds later. (g=9.81m/s二次)
回答 (2)
2007-09-27 3:59 am
✔ 最佳答案
We should now take the top of a building as a reference level~Then applying the equation s = ut + 1/2 at^2
s = (50)(11) + 1/2 (-9.81)(11^2)
= -43.505m
Thus, when taking the ground as reference level, the object is 100 - 43.505 = 56.495m above the ground.
Hope the above information helps =)
By小儒
2007-09-27 3:55 am
u=50, a=9.81, t=11
s=ut+1/2at^2
s= 50x11+1/2 (9.81)(11^2)
s=1143.505m
s>100
so it is on the ground at 11s
s=ut+1/2at^2
s= 50x11+1/2 (9.81)(11^2)
s=1143.505m
s>100
so it is on the ground at 11s
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