# Physics Homework Help Needed ASAP?

A child's water pistol shoots water through a 1.0-mm-diameter hole. If the pistol is fired horizontally 75 cm above the ground, a squirt hits the ground 1.2 m away.

What is the volume flow rate during the squirt? Ignore air resistance.

QQ =

mL/s

Relevance
• falling time t = √2h/g = √1.5/9.806 = 0.391 sec

water speed Vw = x/t = 1.2/0.391 = 3.068 m/sec

let's put dimension in cm

Φ = 1.0 mm = 0.10 cm

Vw = 3.068 m/sec = 306.8 cm/sec

A = 0.7854*Φ^2 = 0.007854 cm^2

Flow Q = A*Vw = 0.007854*306.8 = 2.410 cm^3/sec (ml/sec)

• Login to reply the answers
• The fall time is

t = sqrt(2h/g)

The initial velocity of the water therefore was

v = D/t = D sqrt(g/(2h))

where D is the horizontal distance.

The flow rate is

Q = v A

= A D sqrt(g/(2h))

where A is the area of the opening. If the diameter of the circular opening is d, then A = 1/4 pi d^2.

Q = 1/4 pi d^2 D sqrt(g/(2h))

If I computed correctly upon plugging in the quantities, it gives 2.4 ml/s, but do check it. You will need to convert from m^3/s to cm^3/s.

• Login to reply the answers