- #1

dragon-kazooie

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## 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?*

## Homework Equations

D

_{y}=v

_{i}t + ½ at

^{2}

and

V

_{fy}

^{2}= V

_{iy}

^{2 }+ 2ad

_{y}

## The Attempt at a Solution

First solve for the time in the air: t = d

_{x}/v

_{x}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 = V

_{y}/V

_{x}

Tan 40° = V

_{y}/2m/s

V

_{y}= 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 at

^{2}to be a larger value than v

_{i}t 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?

d

_{y}=v

_{i}t + ½ at

^{2}

d

_{y}=1.68 × 0.75s + ½ -9.8m/s

^{2}× (0.75s)

^{2}

d

_{y}= 1.26 - 2.75

d

_{y}= -1.49m

If I use this formula in blue I get a different final answer which seems more correct:

V

_{fy}

^{2}= V

_{iy}

^{2 }+ 2ad

_{y}

0 = 1.678

^{2}+ 2 × -9.8 × d

_{y}

0 = 2.816 + 2 × -9.8 × d

_{y}

-2.816 = -19.6 d

_{y}

0.143 = d

_{y}

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

**Thank you in advance!**

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