以10m/s沿水平方向抛出一个小球，落地时速度方向与水平方向成60°角，求
（1）小球落地时速度大小？
（2）小球在空间运动的时间？
（3）小球抛出时距水平地面的高度？
（1）cos60°=
=vx v v0 v
则小球落地时的速度大小v=
=v0 cos60°
m/s=20m/s．10 1 2
（2）tan60°=
=vy vx vy v0
则vy=v0tan60°=10
m/s3
又vy=gt
所以小球在空中运行的时间t=
=vy g
s=10 3 10
s．3
（3）h=
gt2=1 2
×10×3m=15m1 2
故小球抛出时距离水平地面的高度为15m．
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