CH.3
A water pistol aimed horizontally projects a stream of water with an initial speed of 5.65 m/s.
(a) How far does the water drop in moving 1.80 m
horizontally?
(b) How far does it travel before dropping a vertical distance of
1.90 cm?
Given that
The horizontal speed, Vx = 5.65 m/s
1)
The horizontal distance, x = 1.80 m
We have a formula for the time as, time(t) = distance / velocity = 0.32 s
We have a formula for the displacement as
s = Vy * t + 0.5at^2
s = 0.5 * 9.8 * (0.32)^2 = 0.50 m
2)
Given that
The vertical distance, y = 1.90 cm = 0.0190 m
To find out the time
y = Vy * t + 0.5at^2
From the above we have
y = 0.5at^2
t = sqrt (2y/a) = sqrt(2*0.0190/9.8) = 0.062 sec
Then the distance is given by the formula
s = Vx * t = 5.65 * 0.062 = 0.350 m
I hope help you !!
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The construction of a water pistol is shown in the figure below. The cylinder with cross-sectional area A, is filled with water and when the piston is pushed (by pulling the trigger), water is forced out the tube with cross-sectional area Ag. The radius of the cylinder and tube are, respectively, 1.30 cm and 1.50 mm, and the center of the tube is a height h = 3.00 cm above the center of the cylinder. (Assume atmospheric pressure is 1.013...
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