枪手沿水平方向对准正前方100m处的靶射击,第一发子弹击中靶上的A点,初速度为500m/s,第二发子弹击中A点正下方5c

lianws2022-10-04 11:39:544条回答

枪手沿水平方向对准正前方100m处的靶射击,第一发子弹击中靶上的A点,初速度为500m/s,第二发子弹击中A点正下方5cm的B点,计算第二发子弹的初速度.

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晚风不吹 共回答了19个问题 | 采纳率89.5%
设出发位置到的A,B垂直距离分别为h和h'
两次初始速度分别为v和v',时间分别为t和t'
t=s/v=100/500=0.2s
h=1/2 gt^2=0.196m
h'=h+0.005=2.001m
h'=1/2 g t'^2,所以,t'=0.64s
v'=156.25m/s
1年前
wqlv 共回答了1个问题 | 采纳率
t=s/v=100/500=0.2s
h=1/2 gt^2=0.196m
h'=h+0.005=2.001m
h'=1/2 g t'^2,所以,t'=0.64s
v'=156.25m/s
1年前
franc001 共回答了1个问题 | 采纳率
t=s/v=100/500=0.2s
h=1/2 gt^2=0.196m
h'=h+0.005=2.001m
h'=1/2 g t'^2,所以,t'=0.64s
是200根号5
1年前
cghcstr 共回答了8个问题 | 采纳率
先要理解为什么两子弹落点不同.
直接原因是竖直方向上下落距离不一样.
对于两个子弹而言.在竖直方向上都是做初速度为0/加速度为g的匀加速直线运动.
唯一不同就是两个子弹下落的时间不同,下落的时间和水平时间是一样的
T1=100/500=0.2秒
T2=100/V
那H1=0.5g(T1的平方)求出H1
H2=H1+0.05...
1年前

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