Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. if their horizontal velocities at the point of ejection are 1 meter/second and 0.5 meters/second respectively, calculate the ratio of their horizontal distances of impact.

 t1 = √2h1/g = √2*0.5/9,8 = 0.319 sec 
t2 = √2h2/g = √2*1.0/9,8 = 0.451 sec 

In which t = times for the vertical movement
h = height
g= gravity (we use standardized measurement of 9.8)

d1 = 1*0.319 = 0.319 m 
d2 = 0.5*0.451 = 0.225 m 

in which d = Horizontal distance

ratio
= di : d2
= 0.319 : 0.225   

 = 3.19 : 2.25

Ke reqd to move from 0 to 50 will be proportional to 50x50 or 2500. this is equal to w.  to change vel from 50 to 150 you need energy prop to 22500 - 2500 = 20,000. this corresponds to 8w.  best answer is a) . work reqd is prop to velocity squared. 

Cis the answer  hope this : d