A kid fires water horizontally from a squirt gun held 1.6 above the ground. It hits another kid 2.1 away square in the back, at a point 0.91 above the ground.

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What was the water's initial speed?

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To calculate the initial speed at which the water is fired from the squirt gun, we can use the principles of projectile motion. The horizontal distance traveled by the water is given as 2.1 m, the vertical distance is 0.91 m, and the initial height from which the water is fired is 1.6 m.

In projectile motion, the horizontal (x) and vertical (y) motions are independent of each other. Let's break down the problem into its components:

1. Horizontal Motion:
The horizontal motion is unaffected by gravity, so the horizontal velocity remains constant throughout the motion. The horizontal distance (x) is given as 2.1 m.

We can use the formula for horizontal motion: distance = velocity × time
Since the initial horizontal velocity is constant, the time of flight will be distance divided by velocity: time = distance / velocity

2. Vertical Motion:
The vertical motion is affected by gravity. We need to determine the initial vertical velocity (Vy) and the time of flight (t) for the vertical motion.

The initial vertical velocity (Vy) is zero because the water is fired horizontally.

We can use the formula for vertical motion: distance = (initial vertical velocity × time) + (0.5 × acceleration due to gravity × time^2)
In this case, the initial vertical velocity is zero, and the distance is 0.91 m. The acceleration due to gravity is approximately 9.8 m/s^2.

Now, let's calculate the values:

1. Calculating the time of flight for horizontal motion:
time = distance / velocity
2.1 m = velocity × time
We need the value of velocity to calculate time, so we'll save this equation for later use.

2. Calculating the time of flight for vertical motion:
0.91 m = (0 × time) + (0.5 × 9.8 m/s^2 × time^2)
0.91 m = 4.9 m/s^2 × time^2
Rearranging the equation to solve for time:
time^2 = 0.91 m / (4.9 m/s^2)
time^2 ≈ 0.186 sec^2
time ≈ √(0.186 sec^2)
time ≈ 0.431 sec

Now that we have the time of flight (t), we can substitute it into the equation from step 1 to find the initial horizontal velocity:

time = distance / velocity
0.431 sec = 2.1 m / velocity
velocity ≈ 2.1 m / 0.431 sec
velocity ≈ 4.868 m/s

Therefore, the initial speed at which the water is fired from the squirt gun is approximately 4.868 m/s.