A hand-pumped water gun is held level at a height of 0.67 m above the ground and

Question

A hand-pumped water gun is held level at a height of 0.67 m 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. A hand-pumped water gun is held level at a height of 0.67 m 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.

Explanation / Answer

The vertical height h = 0.5 gt^2 where t is the time taken to fall vertically through a height h. In this time the distance traveled horizontally = ut = s where u is the horizontal velocity. h = 0.5*g (s/u) ^2 u^2 = 0.5*g s^2/h Gauge pressure P/ density of water ? = 0.5* u^2 = 0.25*g s*s/h Gauge pressure P = 1000*0.25*9.8*7.3*7.3 / 0.67 =194866.42 pascal

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