You have a tank of water sitting on a table and water is squirting out a spigot

Question

You have a tank of water sitting on a table and water is squirting out a spigot on the side. The water is leaving the tank traveling horizontally at a speed of 3.15 m/s. The spigot is a height h=0.910 m above the ground. Well set up the coordinate system to be +x horizontal to the right and +y vertically upward.

What is the initial horizontal component of the velocity just as it leaves the spigot? (How fast is the water traveling horizontally just as it leaves the spigot?)

While the water is in the air, what are the x and y components of the acceleration?
ax:  
ay:

Now lets track the motion of one little chunk of water. How much time elapses between when the water leaves the spigot and when it hits the ground? (Hint: You dont need to worry at all about the horizontal components, and you already have v0y, ay, and y)

Explanation / Answer

Solution: -

initial vertical component of the flow velocity = 0 as mentioned in the question .

so the vertical velocity just as it leaves = 0

2) initial horizontal component of the flow velocity = 3.15 m/s

3) as the water is in air, it is in a state of free fall

so a)y = 9.8 m/s2 downwards

and a) x = 0 because there is no force in horizontal direction on water

4) height h=height , v) x = 3.15 m/s

so chunk of water will exactly behave like projectile motion

now in projectile motion , Max height = given =0.910 m

so in hitting the ground it has to travel this much distance

a)y = -9.8m/s2 , v) y = 0

So, apply equation of motion in Y- direction,

     - 0.910 = -

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