在距地面7.2米高处水平抛出一小球,小球在第1秒内位移为13米,不计空气阻力,取g=10m/s2.小球落地时的速度为_.
在距地面7.2米高处水平抛出一小球,小球在第1秒内位移为13米,不计空气阻力,取g=10m/s2.小球落地时的速度为___.
物理人气:477 ℃时间:2019-10-09 00:17:49
优质解答
第1s内竖直分位移y1=12gt12=12×10×1m=5m,根据平行四边形定则知,第1s内的水平位移x1=132-52=12m,则初速度v0=x1t1=121m/s=12m/s.小球落地时竖直分速度vy=2gh=2×10×7.2m/s=12m/s,则小球落地的速度v=vy2+v02=14...
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