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 5.00 m. a child, who is holding the same gun in a horizontal position, is also sliding down a 45.0° incline at a constant speed of 2.00 m/s. if the child fires the gun when it is 1.00 m above the ground and the water takes 0.329 s to reach the ground, how far will the water travel horizontally?

Refer to the diagrams shown below.

Neglect air resistance.

Case A.
Let t =  the time of flight.
Then for horizontal travel,
t = (5 m)/(u m/s) = 5/u s

For vertical travel,
1.0 m = (1/2)*(9.8 m/s²)*(5/u s)²
5/u = √(1/4.9) = 0.4518
u = 11.0668 m/s

Case B.
There are two contributions to the horizontal velocity
(i) 11.0668 m/s
(ii) (2 m/s)*cos(45) = 1.4142 m/s
The total horizontal velocity is
11.0668 + 1.4142 = 12.481 m/s

Because the time of flight is 0.329 s, the horizontal distance traveled is
d = (12.481 m/s)*(0.329 s) = 4.106 m

The horizontal distance traveled is 4.11 m

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