# Projectile Motion - Niagra Falls

#### crono_

1. The problem statement, all variables and given/known data

Suppose the water at the top of Niagara Falls has a horizontal speed of 2.7 m/s just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a 75° angle below the horizontal?

x direction

a = 0 m/s

v0x = 2.7 m/s

vfx= 2.7 m/s

x = ?

y direction

ay = -9.80 m/s2

v0y = 0 m/s

vfy = ?

y = ?

Other Info

Angle = 75 degrees below horizontal

t = ?

2. Relevant equations

Newton's laws of motion...but I'm not exactly sure which ones as it appears there are two variables missing from each one.

3. The attempt at a solution

I'm studying for my final this Friday. This question plagued me at the beginning of the term, I never got it. The class moved on, and so did I. But I'm looking at it again and am determined to get it, but am still drawing a blank.

I'd like to find t as that would help significantly, but each equation that I've considered with it has another unknown variable as well.

Kind of flustered...

#### ideasrule

Homework Helper
Imagine a water molecule falling off the edge of the waterfall. What's its vertical speed at time t? What's its horizontal speed? How are the vertical & horizontal speeds related to the 75 degrees?

#### ideasrule

Homework Helper
If you do it correctly, you WILL get t.

#### crono_

tan-1 = (vy / vx) = 75 degrees

#### crono_

Wait....

vy / vx = tan 75

vy = tan 75 (vx)

vy = 10.0765 m/s

vfy = voy + at

vfy - voy / a = t

t = 1.0282 s

y = 1/2 (vfy + voy) t

y = 1/2 (10.0765 m/s + 0 m/s) 1.0282 s

y = 5.18 m ----> 5.2 m with sig figs

Hrm....interesting.................

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