A water pistol aimed horizontally projects a stream of water with an initial speed of 7.20 m/s. How far does it travel before dropping a vertical distance of 1.95 cm?
(horizontal velocity)* (time to fall 1.95 cm)
8.25m
To find the distance traveled by the water stream before dropping a vertical distance, we can use the equations of projectile motion.
First, let's break down the information given:
Initial speed of the water stream (horizontal component) = 7.20 m/s
Vertical distance dropped = 1.95 cm
We need to find the horizontal distance traveled by the water stream before dropping 1.95 cm vertically.
In projectile motion, we can split the motion into horizontal and vertical components. The motion in the horizontal direction is at a constant velocity, while the motion in the vertical direction is influenced by gravity.
The time taken for the water stream to drop vertically can be found using the equation:
Δy = v₀y * t + (1/2) * g * t²
Where:
Δy is the vertical distance dropped,
v₀y is the vertical component of the initial velocity,
t is the time taken for the motion,
and g is the acceleration due to gravity (approximately 9.8 m/s²).
Converting the vertical distance to meters:
Δy = 1.95 cm = 0.0195 m
Using the equation:
0.0195 m = 0 * t + (1/2) * 9.8 m/s² * t²
0.0195 m = (4.9) * t²
Simplifying the equation:
t² = 0.0195 / 4.9
t² = 0.00398
Taking the square root of both sides:
t = √0.00398
t ≈ 0.063 s
Now that we have the time, we can find the horizontal distance using the equation:
Δx = v₀x * t
Where:
Δx is the horizontal distance traveled,
v₀x is the horizontal component of the initial velocity,
and t is the time taken.
The horizontal component of the initial velocity is the same as the initial speed since there is no horizontal acceleration. Therefore:
v₀x = 7.20 m/s
Using the equation:
Δx = 7.20 m/s * 0.063 s
Δx ≈ 0.454 m
Therefore, the water stream travels approximately 0.454 meters before dropping a vertical distance of 1.95 cm.