In a friendly neighborhood squirt gun contest a participant runs at 7.8 m/s horizontally off the back deck and fires her squirt gun in the plane of her motion but 45' above horizontal. The gun can shoot water at 11 m/s relative to the barrel, and she fires the gun O.42 s after leaving the deck. (a) What is the initial velocity of the water particles as seen by an observer on the ground? Give your answer in terms of the horizontal and vertical components. O) At the instant she fires, the gun is 1.9 m above the level ground. How far will the water travel horizontally before landing?
first the shooter:
horizontal velocity = us = 7.8 m/s forever
vertical velocity = vs = - g t =-9.81 t
at .42 s this is - 9.81*.42 = -4.12 m/s
now the water
horizontal velocity = u= 7.8 + 11*.707
= 15.6 m/s forever
vertical velocity =vw = -4.12 + 11*.707
= 3.66 AT .42 s
use u = 15.6 and v = 3.66 to answer part (a)
now
vw = 3.66 - 9.81(t-.42)
uw = 15.6
Hiw = initial height of water at .42 = 1.9
so create new t for time after .42 and water particle doing that:
vw = 3.66 - 9.81 t
uw = 15.6
Hiw = initial height of water at .42 = 1.9
initial x of water = .42 * 7.8=3.28
(problem does not say if interested in how far from takeoff or how far after firing)
how long in air after firing?
0 = 1.9 + 3.66 t -4.9 t^2
solve quadratic I get t = 1.1 second
so horizontal distance after discharge = 15.6 *1.1 = 17.2 meters
If you are supposed to include the horizontal 3.28 meters the gun traveledd before discharge then
17.2 +3.3 = 20.5 meters
Thank you very much Damon.
You are welcome.
To find the initial velocity of the water particles as seen by an observer on the ground, we need to consider the vector components of the participant's velocity and the gun's water velocity.
Let's break down the participant's velocity into horizontal (Vx) and vertical (Vy) components. The horizontal component remains constant, which is 7.8 m/s, as there is no horizontal acceleration involved.
The vertical component of the participant's velocity (Vy) can be determined by multiplying the initial velocity (7.8 m/s) by the sine of the angle of 45' above horizontal:
Vy = 7.8 m/s * sin(45') = 7.8 m/s * 0.7071 = 5.51 m/s
Now let's consider the water velocity relative to the barrel. The given water velocity is 11 m/s, which represents the magnitude of the water velocity. However, we need to break it down into horizontal and vertical components as well.
The horizontal component of the water velocity (Vx_water) is equal to the participant's horizontal velocity (Vx) because they are firing the gun in the plane of their motion. So, Vx_water = Vx = 7.8 m/s.
The vertical component of the water velocity (Vy_water) is equal to the participant's vertical velocity (Vy) plus the water's velocity relative to the barrel. Therefore:
Vy_water = Vy + 11 m/s
Vy_water = 5.51 m/s + 11 m/s = 16.51 m/s
Now we have the horizontal and vertical components of the initial velocity of the water particles as seen by an observer on the ground:
Horizontal component: Vx_water = 7.8 m/s
Vertical component: Vy_water = 16.51 m/s
To determine the horizontal distance the water travels before landing, we can use the horizontal component of the water's initial velocity and the time it takes for the water to land.
Given that the water is fired 0.42 s after the participant leaves the deck and the vertical distance between the gun and the ground is 1.9 m, we can use the equation:
Horizontal distance = Vx_water * time
Substituting the values:
Horizontal distance = 7.8 m/s * 0.42 s = 3.28 m
Therefore, the water will travel horizontally for approximately 3.28 meters before landing.