# NPY and shot （物理题）

NPY is going to have a PE test.One of the test subjects is throwing the shot.The height of NPY is H meters.He can throw the shot at the speed of v0 m/s and at the height of exactly H meters.He wonders if he throws the shot at the best angle,how far can he throw ?(The acceleration of gravity, g, is 9.8m/s29.8m/s2)

InputThe first line contains a integer T(T10000)T(T≤10000),which indicates the number of test cases.
The next T lines,each contains 2 integers H(0h10000m)H(0≤h≤10000m),which means the height of NPY,and v0(0v010000m/s)v0(0≤v0≤10000m/s), which means the initial velocity.OutputFor each query,print a real number X that was rounded to 2 digits after decimal point in a separate line.X indicates the farthest distance he can throw.Sample Input

2
0 1
1 2

Sample Output

0.10
0.99



Hint

If the height of NPY is 0,and he throws the shot at the 45° angle, he can throw farthest.


#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstdio>
using namespace std;
int h,v;
double judge(double a){
//v*cosa(vsina+sqrt(v2sin2a+2gh)/g
return (v*cos(a))*((v*sin(a)+sqrt(v*v*sin(a)*sin(a)+2*9.8*h))/9.8);
}
int main(){

int t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&h,&v);
double l=0,r=0.5*3.1415926535;
double mid1,mid2;
while(r-l>1e-8){
mid1=(l+r)/2.0;
mid2=(mid1+r)/2.0;
if(judge(mid1)>judge(mid2)){
r=mid2;
}else{
l=mid1;
}
}
printf("%.2f\n",judge(r));
}

}