(1)由机械能守恒定律得,A点时的机械能 E A = E B =
1
2 m
v 2B +mgh
代入数据解得,E A=68J
(2)根据机械能守恒得:
mg(H-h)=
1
2 m
v 2B -
1
2 m
v 20
代入可解得v 0=6m/s
(3)L=v 0•t 1
t 1 =
2(H-h)
g =
v y
g =0.8s
综合可解得L=4.8m
答:
(1)小球从A点抛出时的机械能为68J.
(2)小球从A点抛出时的初速度v 0为6m/s.
(3)B点离竖直墙壁的水平距离L为4.8m.
(1)由机械能守恒定律得,A点时的机械能 E A = E B =
1
2 m
v 2B +mgh
代入数据解得,E A=68J
(2)根据机械能守恒得:
mg(H-h)=
1
2 m
v 2B -
1
2 m
v 20
代入可解得v 0=6m/s
(3)L=v 0•t 1
t 1 =
2(H-h)
g =
v y
g =0.8s
综合可解得L=4.8m
答:
(1)小球从A点抛出时的机械能为68J.
(2)小球从A点抛出时的初速度v 0为6m/s.
(3)B点离竖直墙壁的水平距离L为4.8m.