A solid sphere with a diameter of 0.16 m is released from rest; it then rolls without slipping down a ramp,dropping through a vertical height of h1 = 0.66 m.The ball leaves the bottom of the ramp,which is h2 = 1.39 m above the floor,moving horizontal

A solid sphere with a diameter of 0.16 m is released from rest; it then rolls without slipping down a ramp,dropping through a vertical height of h1 = 0.66 m.The ball leaves the bottom of the ramp,which is h2 = 1.39 m above the floor,moving horizontal
A solid sphere with a diameter of 0.16 m is released from rest; it then rolls without slipping down a ramp,dropping through a vertical height of h1 = 0.66 m.The ball leaves the bottom of the ramp,which is h2 = 1.39 m above the floor,moving horizontally.
(a) Through what horizontal distance d does the ball travel before landing?
1.62 m
(b) How many revolutions does the ball make during its fall?
第一步我做对了,第二步算了好几遍都是错的,最好有步骤让我知道我错在哪里

A solid sphere with a diameter of 0.16 m is released from rest; it then rolls without slipping down a ramp,dropping through a vertical height of h1 = 0.66 m.The ball leaves the bottom of the ramp,which is h2 = 1.39 m above the floor,moving horizontal
球绕直径的转动惯量：J=2/5mR²
根据无滑动条件,转速与前进速度的关系：ω=v/R
根据能量守恒,落到坡底时势能转化为平动动能和转动动能,由此求出水平抛出时的角速度：1/2mv²+1/2Jω²=mg*h1
球平抛后落地的时间：1/2gt²=h2
在这段时间里球的运动圈数：2πn=ωt
代入：R=0.08,h1=0.66,h2=1.39,g=9.8
解得：n=3.22

把你算的步骤列出来看看。

你把它翻译过来 可能很多会物理的就能帮你解答了