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2 years ago

Water at the top of Horseshoe Falls (part of Niagara Falls)

Physics

1 answer:

ale4655 [162]2 years ago

8 0

Answer:

$\theta= (-74.42)^{\circ} C$

Explanation:

Horizontal speed of water, $v_{xf}=9\ m/s$

Height, h = -53 (below pool)

We can find firstly the final vertical speed of the water using third equation of kinematics. So

$v^2_{yf}=u^2_{yi}+2(-g)h\\\\v^2_{yf}=2\times -9.8\times -53\\\\v_{yf}=32.23\ m/s$

Let $\theta$ is the angle where the falling water moving as it enters the pool. So,

$\tan\theta=\dfrac{v_{yf}}{v_{xf}}\\\\=\dfrac{-32.3}{9}\\\\=-74.42^{\circ} C$

Hence, the angle is (-74.42)°C.

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Explanation:

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