Question

Consider a horizontally oriented syringe containing water located at a height of 1.25 m above the ground. The diameter of the plunger is 8 mm and the diameter of the nozzle is 2 mm.The plunger is pushed with a constant speed of 0.25 m/s. Find the horizontal range of water stream on the ground.

Answer 1

the answer is 2m

•    A₁V₁ = A₂V₂

• ⇒ A₁V₁/A₂ = ( πr₁² / πr₁² ) * V₂

• ⇒ V₂ = (D/d)² V₁ = (8×10⁻³/2×10⁻³)² * 0.25m/s

• ⇒ V₂ = 4 m/s horizontal

Vertical component of the velocity is zero

Now,

•     H = (1/2)gt²

• ⇒  t = √(2H/g)

Range is given by

• R = V₂t  = 4 * √(2H/g) = 4 * [( 2 * 1.25 ) / 10 ] = 2m