A ball is thrown vertically upwards from the ground . It hits the ground 5 seconds later .
(a) Find the initial velocity
(b) Find the final velocity
(c) Find the maximum height reached by the ball
phy~~~~~
2007-03-07 6:53 am
回答 (2)
2007-03-07 7:26 am
✔ 最佳答案
a)v=u+at
0=u+(-10)(2.5)
u=25m/s
b)
v=u+at
v=0+(10)(2.5)
v=25m/s
c)
ss=ut+att
ss=25+(10)(2.5)(2.5)
s=6.12m
d正負自己攪返
2007-03-07 10:16 am
we first take downward motion as positive
and the acceleration due to gravity as 9.81 m/s
(a) by symmetry, time t to reach the maximum height is 5/2=2.5 s
by v=u+gt
0=u+(9.81)(2.5)
0=u+24.5
so u=-24.5 m/s (or 24.5 m/s upward)
(b) by symmetry, time t to reach the ground is 5/2=2.5 s
by v=u+gt
v=0+(9.81)(2.5)
so v=24.5 m/s
another analysis:
by conservation of energy
the initial K.E. totally changed to P.E. at first
and then release back to K.E. when the ball return to the ground level
so the magnitude of velocity should be the same as inititial value
so it's 24.5 m/s and downward
(c) by conservation of energy
P.E. gain=K.E. loss
mgh=(1/2)mu^2
gh=(1/2)u^2
(9.81)h=(1/2)(24.5)^2
so h=30.7 m
and the acceleration due to gravity as 9.81 m/s
(a) by symmetry, time t to reach the maximum height is 5/2=2.5 s
by v=u+gt
0=u+(9.81)(2.5)
0=u+24.5
so u=-24.5 m/s (or 24.5 m/s upward)
(b) by symmetry, time t to reach the ground is 5/2=2.5 s
by v=u+gt
v=0+(9.81)(2.5)
so v=24.5 m/s
another analysis:
by conservation of energy
the initial K.E. totally changed to P.E. at first
and then release back to K.E. when the ball return to the ground level
so the magnitude of velocity should be the same as inititial value
so it's 24.5 m/s and downward
(c) by conservation of energy
P.E. gain=K.E. loss
mgh=(1/2)mu^2
gh=(1/2)u^2
(9.81)h=(1/2)(24.5)^2
so h=30.7 m
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