A water pistol aimed horizontally projects a stream of water with an initial speed of 5.65 m/s. How far does it travel before dropping a vertical distance of 1.50 cm?
Multiply the horizontal speed component (which remains constant) by the time T that it takes the water to fall 1.50 cm.
(g/2) T^2 = 0.015 m
T = sqrt(0.03/g) = 0.0553 s
X = 0.313 m not very far!
To find the horizontal distance traveled by the stream of water before it drops a vertical distance of 1.50 cm, we can use the equations of motion.
The horizontal motion is not affected by gravity, so the initial horizontal velocity remains constant.
Given:
Initial speed of the water stream, u = 5.65 m/s
Vertical distance dropped, h = 1.50 cm = 0.015 m
First, let's find the time it takes for the water to drop 0.015 m vertically.
Using the equation of motion for vertical displacement:
h = ut + (1/2)gt^2
where:
h = vertical displacement
u = initial vertical velocity (0 since it drops)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Rearranging the equation, we have:
0.015 = 0 + (1/2)(9.8)(t^2)
0.015 = 4.9t^2
t^2 = 0.015 / 4.9
t^2 ≈ 0.003061
t ≈ √0.003061
t ≈ 0.0554 seconds
Now, we know that the time taken for the water to drop vertically is approximately 0.0554 seconds.
To find the horizontal distance traveled, we'll use the formula for horizontal distance:
Horizontal distance = horizontal velocity * time
The horizontal velocity remains constant and is equal to the initial speed of the water stream, which is 5.65 m/s.
Horizontal distance = 5.65 m/s * 0.0554 s
Horizontal distance ≈ 0.313 meters
Therefore, the water stream will travel approximately 0.313 meters horizontally before dropping a vertical distance of 1.50 cm.