Question

we ( A_{1} v_{1}=A_{2} V_{2} )
( Rightarrow 8^{2} times 0.25=2^{2} times v_{2} )
( Rightarrow frac{64 times 1}{4}=442 )
a) ( quad V_{2}=4 mathrm{m} / mathrm{s} )
Solved ision i ( t=sqrt{frac{24}{g}}=sqrt{frac{2 times 1.25}{10}}=sqrt{25} )
Range: ( 4 times .5 )
( 2 m )

# Consider a horizontally oriented syringe containing water located at a height of 1.25 m above the cround. 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. Take g = 10 m/s [IIT-JEE 2004] D 8mm 14- 11.25m Ground

Solution