Question

above the ground and fired. The water stream from the gun hits the ground a horizontal distance of 7.3 m from the muzzle. Find the gauge pressure of the water gun’s reservoir at the instant when the gun is fired. Assume that the speed of the water in the reservoir is zero and that the water flow is steady. Ignore both air resistance and the height difference between the reservoir and the muzzle.

Answer 1

`P_g=1.1373Mpa`

Step-by-step explanation:

A hand-pumped water gun is held level at a height of 0.75 m, this part miss of the question so now:

h=0.75 m

x=7.3 m

`h=(1)/(2)*a*t^2`

`a=g`

`t=\sqrt{(2*h)/(g)}=\sqrt{(2*0.75m)/(9.8m/s^2)}`

`t=0.153s`

`v=(x)/(t)=(7.3m)/(0.153s)=47.6m/s`

Gauge Pressure can be find

`P_g=(1)/(2)*1000*v^2`

`P_g=(1)/(2)*1000*(47.69m)^2`

`P_g=1.1373Mpa`

Answer 2

The Gauge pressure ( Pg ) of the water gun's reservoir at the Instant when the gun is fired is = 175219.2 Pa

Given data :

Horizontal distance travelled by the water from gun ( x ) = 7.3 m

( Missing Information ) Vertical distance of water gun ( h ) = 0.75 m

Next : determine the value of the Gauge pressure ( Pg )

Pg = 1/2 * 1000 * v² --- ( 1 )

where; v = x / t ----- ( 2 )

t =
`√(2*h ) / g )` =
`√((2* 0.75)/9.81 )` = 0.39 sec

back to equation 2

v = 7.3 / 0.39 = 18.72 m/s

back to equation ( 1 )

Pg = 1/2 * 1000 * ( 18.72)²

= 175219.2 Pa

Hence we can conclude that the gauge pressure of the water gun's reservoir is 175219.2 Pa