# Kinematic equation problem -- A fountain shoots water out of the ground ....

dragon-kazooie

## Homework Statement

1. A fountain shoots water out of the ground at a 40° angle. It’s horizontal velocity is 2 m/s and the water lands 3 meters away. What is the maximum height that the water reaches during that time?

Dy =vit + ½ at2

and

## The Attempt at a Solution

First solve for the time in the air: t = dx /vx t = 3m/2m/s = 1.5s
Divide that in half to find the time at maximum height: t = 0.75s

Solve for initial vertical velocity:
Tan θ = Vy /Vx
Tan 40° = Vy /2m/s
Vy = 1.68 m/s

Now I get two different answers when I use different equations, and I don't see why. Using this one that I did in purple must be wrong, because it doesn't make sense for ½ at2 to be a larger value than vit and I get a negative number which can't be right. But I don't see where the mistake is! What am I doing wrong?

dy =vit + ½ at2
dy =1.68 × 0.75s + ½ -9.8m/s2 × (0.75s)2
dy = 1.26 - 2.75
dy = -1.49m

If I use this formula in blue I get a different final answer which seems more correct:
0 = 1.6782 + 2 × -9.8 × dy
0 = 2.816 + 2 × -9.8 × dy
-2.816 = -19.6 dy
0.143 = dy

Is the one in blue correct? What did I do wrong with the one in purple?

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Homework Helper
Gold Member
2022 Award
@dragon-kazooie , your first method is correct. You got a crazy answer because the given conditions are impossible. There is no way that a jet of only 2m/s can reach 3m horizontally, even at 45 degrees.
Your second method appeared to work because you do not need the 3m range information, and you effectively ignored it.

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• dragon-kazooie
dragon-kazooie
Thank you @haruspex! That was driving me crazy!