We shall apply Torricelli's law here
It states that the velocity of fluid leaving the container from hole will be same as that of particle dropped form height of the hole (height from the top of container)
For the hole made from three quarter height from ground will have height from the surface h/4 and velocity v1=sqrt{2×(h/4)×g}
Let the range be R1
For the hole made from half the depth of tank will have height from the surface h/2 and velocity v2=sqrt{2×(h/2)×g}
Let the range be R2
Now the water ejected out has horizontal velocity and the range can be calculated by
R=v.t
whre t is the time taken by jet to reach ground and v is horizontal velocity
Now the initial component of vertical velocity is zero so t=sqrt(2H/g)
Where H is height above the ground
Respective time for stream for both holes will be
T1=sqrt{2×(3h/4 )/g}
T2=sqrt{2×(h/2 )/g}
So R1=v1×T1
=sqrt{2×(h/4)×g}×sqrt{2×(3h/4 )/g}
=√3.h
R2=v2×T2
=sqrt{2×(h/2)×g}×sqrt{2×(h/2)/g}
=h
So difference is
R1-R2=(√3-1)h
=(1.73-1)h
=.0.73h
