Projectile Motion - Niagra Falls

[crono_]

Homework Statement

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

v_0x = 2.7 m/s

v_fx= 2.7 m/s

x = ?

y direction

a_y = -9.80 m/s²

v_0y = 0 m/s

v_fy = ?

y = ?

Other Info

Angle = 75 degrees below horizontal

t = ?

Homework 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.

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]

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]

If you do it correctly, you WILL get t.

 

[crono_]

tan^-1 = (v_y / v_x) = 75 degrees

 

[crono_]

Wait...

v_y / v_x = tan 75

v_y = tan 75 (v_x)

v_y = 10.0765 m/s



v_fy = v_oy + at

v_fy - v_oy / a = t

t = 1.0282 s



y = 1/2 (v_fy + v_oy) 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....