英语翻译A hockey ball is ‘flicked’ off the ground with initial v

1）draw a parabola which stands for the way the hocket ball passed,conect the three ponits as an isoceles triangle.画一个抛物线,将三点（顶点、起点、终点）连接为一个三角形.
Suppose the whole distant between the start position and the final point is S,and the horizontal distance could be seperated by the highest point,then we could get S=S1+S2.S1 stands for the first period horizontal distance,and S2 stand for the remaining distance.
假设起点终点间距为S,则水平距离应可以按顶点拆分为S1和S2,那么S=S1+S2
2) in the begining place,the ball's initial velocities is V(Vy=2.7 m s-1 upwards,and Vx=11.4m s-1 horizontal) 在开始阶段,球的初始速度如已知条件所示.
in the top place,the uowards velocity should be 0,while the height it reached is H.在顶部 垂直速度变为0,而此时高度设为H.
3) According to conservation law of energy,in the upwards direction,you could get "m * Vy * Vy = m * g * H" 按照能量守恒定理 可以得到以下恒等式,即质量与垂直速度平方值的和 等于 势能的增加.
Then you could get:H = Vy * Vy / g = 7.29 / 9.8 = 0.744 meters
由此计算出球所能达到的最高高度为0.744米.
4) According to the triangle' length proportion,H :S1 = 2.7 :11.4
so S1 = 11.4 x 0.744 / 2.73 = 3.14 meters
按照三角形边长比例 得到水平与垂直比 应与 初始速度的水平垂直分量比一致,得到S1=3.14米.
5) Since H = 1/2 *( g * t * t ) you could get the time from highest to the lowest point.t = 0.39 second
Since S2 = Vx * t = 11.4 x 0.39 = 4.446 meters.
S =S1 + S2 = 3.14 + 4.446 = 7.586 meters
我也是一时兴起,大概推算的,要是有问题的话,应该在第4点那里.先给你参考吧.
楼上的 snakeslave 的回答应该是对的.刚才没仔细看,

就是一个斜抛运动吗 球在初始时刻有水平速度Vx=11.4m/s 竖直速度Vy=2.7m/s 且从地面抛出 设落地时经过的时间为t 竖直位移为Sy,水平位移为Sx
当落地时 竖直位移为零 即
Sy=Vy*t-0.5*g*t^2=0 (1)
水平方向有
Sx=Vx*t (...

lack of necessary conditions like force, time

who is good at physics here?
a ball given the upward v and horizontal v, try to calculate the distance, have no idea how to do that.