Consider a horizontally oriented syringe containing water locate at a height of `1.25 m` above the ground. Thediameter of the plunger is `8 mm` and the diameter of the nozzle is `2 mm`. The plunger is pushed with a constant speed of `0.25m//s.` Find the horizontal range of water stream on the grond. Take `g=10 m//s^(2)`

Explanation

From equation of continuity
`A_(1)v_(1)=A_(2)v_(1)`
`implies(A_(1))/(A_(2)) v_1=((pir_(1)^(2))/(pir_(2)^(2)))v_(1)`
or `v_(2)=(D/d)^(2)v_(1)=((8xx10^(-3))/(2xx10^(-3)))^(2)xx0.25 m//s`
`=4m//s` (horizontal)
Vertical component of the velocity is zero
Now, `H=1/2"gt"^(2)`
`implies t sqrt((2H)/g)`
Range is given by `R=v_(2)t=v_(2)`
`sqrt((2H)/g)=4xxsqrt((2xx1.25)/10)=2m`