A water pistol aimed horizontally projects a stream of water with an initial speed of 5.00 m/s.
(a) How far does the water drop in moving 1.00 m hori-
zontally?
(b) How far does it travel before dropping a vertical
distance of 1.00 cm?
(a) It takes
t = 1.00/5.00 = 0.2 seconds
to travel 1.0 meter horizontally.
Use the formula
(elevation change) = (g/2) t^2
(b) To fall 1.00 cm, the time required is
given by
0.01 = (g/2) t^2
t = 0.0452 s
The horizontal distance traveled by the water in that time is
5*0.0452 meters
.2 seconds
.226
To solve this problem, we can use the equations of motion for both horizontal and vertical motion.
Let's start with part (a) of the question:
(a) How far does the water drop in moving 1.00 m horizontally?
We know the initial horizontal velocity (Vx) is 5.00 m/s. The time it takes for the water to move horizontally (t) is given by the equation:
t = distance / velocity
Substituting the values, we have:
t = 1.00 m / 5.00 m/s
t = 0.20 s
Since the water moves horizontally at a constant velocity, the vertical distance dropped (y) can be calculated using the equation:
y = 0.5 * g * t^2
where g is the acceleration due to gravity and is approximately equal to 9.8 m/s^2.
Substituting the values, we have:
y = 0.5 * 9.8 m/s^2 * (0.20 s)^2
y = 0.196 m
Therefore, the water drops 0.196 meters in moving 1.00 meter horizontally.
Now let's move on to part (b) of the question:
(b) How far does it travel before dropping a vertical distance of 1.00 cm?
In this case, we need to find the horizontal distance traveled (x) before the water drops vertically by 1.00 cm.
Using the equation of horizontal motion:
x = Vx * t
Substituting the values, we have:
x = 5.00 m/s * 0.20 s
x = 1.00 m
Therefore, the water travels 1.00 meter before dropping a vertical distance of 1.00 cm.
To find the distance the water drops in moving horizontally, we can use the formula for horizontal distance traveled by an object in projectile motion. The formula is:
D = Vx * t
Where:
D is the horizontal distance traveled,
Vx is the horizontal component of velocity,
t is the time taken.
(a) To find the distance the water drops in moving 1.00 m horizontally, we need to determine the time it takes for the water to cross this distance. Since the horizontal component of velocity remains constant, we can use the formula:
t = D / Vx
Given that the initial speed is 5.00 m/s, the horizontal component of velocity is also 5.00 m/s (since the water pistol is aimed horizontally). Substituting the values into the formula:
t = 1.00 m / 5.00 m/s
t = 0.20 s
Now we can calculate the distance the water drops by substituting the obtained time value into the formula:
D = Vx * t
D = 5.00 m/s * 0.20 s
D = 1.00 m
Therefore, the water drops a distance of 1.00 m in moving 1.00 m horizontally.
(b) To find the distance traveled before dropping a vertical distance of 1.00 cm, we need to consider the vertical motion of the water droplet. We can use the formula for vertical distance traveled in projectile motion:
Dy = Vy * t + 0.5 * g * t^2
Where:
Dy is the vertical distance traveled,
Vy is the initial vertical component of velocity,
t is the time taken,
g is the acceleration due to gravity.
In this case, the initial vertical velocity of the water droplet is 0 m/s (since it is moving horizontally), and we can assume the acceleration due to gravity to be -9.8 m/s^2 (negative because it's in the downward direction). Substituting these values into the formula:
Dy = 0 * t + 0.5 * (-9.8 m/s^2) * t^2
Dy = -4.9 * t^2
We know that the vertical distance dropped is 1.00 cm, which is equal to 0.01 m. Replacing Dy with 0.01 m:
0.01 m = -4.9 * t^2
Solving for t:
t^2 = (0.01 m) / (-4.9)
t^2 = -0.002040816
Since time can't be negative, we discard the negative value. Taking the square root of the positive value:
t ≈ 0.045 s
Now we can calculate the horizontal distance traveled by the water droplet using the time obtained:
D = Vx * t
D = 5.00 m/s * 0.045 s
D ≈ 0.225 m
Therefore, the water travels approximately 0.225 m before dropping a vertical distance of 1.00 cm.