（1）物体从一定的高度自由下落,若将下落总时间分为两段相等的时间,则下落高度之比是多少?
（2）一辆汽车从静止开始作匀加速直线运动.从开始运动到驶过100m标记时,速度增加到10m/s,在驶过200m标记时,速度将增加到多少?
（3）楼上落下一个玻璃球,经过窗口上沿时速度为6.0m/s,离开窗口下沿时速度为8.0m/s ,问窗口的高度是多少?
题型：【】 标签：物理
题目：
答案及解析
(1)
H = (gt^2)/2
H1 = [g*(t/2)^2]/2 = H/4
H2 = H - H1 = H*(3/4)
H1:H2 = 1:3
(2)
v1 = at . t = v1/a
s1 = (at^2)/2 = (v1)^2/(2a) = 100/(2a) = 100
a = 0.5m/(s^2)
S2 = (at^2)/2 = (t^2)/4 = 200 . t = 20√2 s
v2 = at = 0.5*20√2 = 14.142 m/s
(3)
v1 = g*t1 = 6.0m/s . t1 = 6/g
H1 = g*(t1)^2 / 2 = 18/g (m)
H2 = g*(t2)^2 / 2 = 32/g (m)
H = H2 - H1 = 14/g (m)
取 g = 10 m/(s^2) 则 H = 1.4m .窗户很矮呀,不对吧?
H = (gt^2)/2
H1 = [g*(t/2)^2]/2 = H/4
H2 = H - H1 = H*(3/4)
H1:H2 = 1:3
(2)
v1 = at . t = v1/a
s1 = (at^2)/2 = (v1)^2/(2a) = 100/(2a) = 100
a = 0.5m/(s^2)
S2 = (at^2)/2 = (t^2)/4 = 200 . t = 20√2 s
v2 = at = 0.5*20√2 = 14.142 m/s
(3)
v1 = g*t1 = 6.0m/s . t1 = 6/g
H1 = g*(t1)^2 / 2 = 18/g (m)
H2 = g*(t2)^2 / 2 = 32/g (m)
H = H2 - H1 = 14/g (m)
取 g = 10 m/(s^2) 则 H = 1.4m .窗户很矮呀,不对吧?