Question

某人一处速度v0=2m/s将质量为m的小球抛出，小球落地式的速度为4m/s，求：小球刚被抛出时离地面的高度。

Answer 1

动能定理：mgh=1/2mvt^2-1/2mvo^2
代入数据得：10h=0.5*16-0.5*4
h=0.6m
希望采纳

Answer 2

mgh=1/2mvt^2-1/2mvo^2
h=(1/2mvt^2-1/2mvo^2)/mg=(1/2m*4^2-1/2m*2^2)/10m=0.6