When a water gun is fired while being held horizontally at a height of 1.00 m above ground level, the water travels a horizontal distance of 4.80 m. A child, who is holding the same gun in a horizontal position, is also sliding down a 45° incline at a constant speed of 1.64 m/s. If the child fires the gun when it is 1.1 m above the ground and the water takes 0.3700 s to reach the ground, how far will the water travel horizontally?
To find the horizontal distance traveled by the water, we can use the equation:
d = v * t
where:
d is the horizontal distance traveled by the water,
v is the horizontal velocity of the water, and
t is the time it takes for the water to reach the ground.
We know that the velocity of the water is equal to the horizontal component of the child's velocity since the child is holding the water gun horizontally. Therefore, we need to calculate the horizontal component of the child's velocity.
Given:
Height of the gun above the ground (h) = 1.1 m
Time taken for the water to reach the ground (t) = 0.3700 s
Velocity of the child sliding down the incline (v_child) = 1.64 m/s
To find the horizontal component of the child's velocity, we can use trigonometry. The horizontal component can be calculated using the formula:
v_x = v_child * cos(θ)
where θ is the angle of the incline.
Since the child is sliding down a 45° incline, we have:
θ = 45°
Now, we can calculate the horizontal component of the child's velocity:
v_x = 1.64 m/s * cos(45°)
v_x = 1.64 m/s * (sqrt(2)/2)
v_x = 1.64 m/s * 0.7071
v_x ≈ 1.1594 m/s
Now that we have the horizontal component of the child's velocity, we can calculate the horizontal distance traveled by the water:
d = v_x * t
d = 1.1594 m/s * 0.3700 s
d ≈ 0.4288 m
Therefore, the water will travel approximately 0.4288 meters horizontally.
To find the horizontal distance traveled by the water, we need to consider the projectile motion of the water.
First, let's find the initial horizontal velocity of the water. Since the child is sliding down the incline at a constant speed of 1.64 m/s, the horizontal component of this velocity will also be 1.64 m/s (because there is no acceleration in the horizontal direction).
Next, we need to determine the time it takes for the water to reach the ground. Given that the water takes 0.3700 s to reach the ground, we'll use this value for calculations.
Since the water is fired when it is 1.1 m above the ground, let's find how long it will take for the water to fall this distance.
Using the equation for vertical displacement:
y = ut + (1/2)gt^2
where:
y = vertical displacement (1.1 m)
u = initial vertical velocity (0 m/s, as the water is not moving vertically initially)
g = acceleration due to gravity (-9.8 m/s^2)
t = time (unknown)
Substituting the known values and solving for t:
1.1 = 0 + (1/2)(-9.8)t^2
(1/2)(-9.8)t^2 = -1.1
4.9t^2 = 1.1
t^2 = 1.1/4.9
t ≈ 0.3364 s (rounded to four decimal places)
Since it takes the water 0.3700 s to reach the ground, this means the water traveled an additional time of t = 0.3700 - 0.3364 ≈ 0.0336 s (rounded to four decimal places).
Now, we can find the horizontal distance covered by the water during this additional time by multiplying the horizontal velocity by time:
Distance = velocity * time
Distance = 1.64 m/s * 0.0336 s
Distance ≈ 0.0551 m (rounded to four decimal places)
Therefore, the water will travel an additional horizontal distance of approximately 0.0551 meters.