A ball with mass 10g is thrown vertically upwards at 20ms^-1 . ignoring the air resistance and taking g=10ms^-2 , calculate
A)how high it goes,
B)the time taken to reach this height
C)the time taken to return to its starting point,
D)how much PE is gained when it reaches the highest point,
E)at what speed does it reach its starting point?
希望有式解答...
地心吸力問題...(計數的)
2010-01-04 2:44 am
回答 (1)
2010-01-04 4:03 am
✔ 最佳答案
(a) Use v^2 = u^2 + 2aswith u = 20 m/s, a = -g = -10 m/s2, v = 0 m/s, s =?
0 = 20^2 + 2(-10)s
s = 20 m
(b) Use v = u + at
with u = 20 m/s, v = 0 m/s, a = -10 m/s2, t = ?
0 = 20 + (-10)t
t = 2 s
(c) By symetry, time return to starting point = 2 x 2 s = 4 s
(d) PE gain = 10/1000 x 10 x 20 J = 2 J
(e) By symetry, speed = initial speed of projection = 20 m/s
or you could use energy conservation,
(1/2)(10/1000)v^2 = 2
hence, v = 20 m/s
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