Question

Water at the top of Horseshoe Falls (part of Niagara Falls) is moving horizontally at 9.2 m/s as it goes off the edge and plunges 53 m to the pool below.
Part A
If you ignore air resistance, at what angle with respect to the vertical is the falling water moving as it enters the pool?

Answer 1

Water at the top of Horseshoe Falls (part of Niagara Falls) is moving horizontally at 9.2 m/s, the angle with respect to the vertical  is mathematically given as

$\theta=15.93$

What angle with respect to the vertical is the falling water moving as it enters the pool?

Generally, the equation for velocity is mathematically given as

V^2=u^2+2as

Therefore

V^{2}=0+2(9.81)(53)

V=32.2m/s

In conclusion, The angle

$\theta =tan^{-1}\frac{(9.20m/s)}{(32.23m/s)}$

$\theta=15.93$

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