We use the formula to find the below formula to find the time, t

h = \frac{1}{2}gt^{2}

We have the height h = 427 m, g = 9.8 \frac{m}{s^{2}}. Plugging these in and solving for t, gives us,

t^{2} = \frac{427 m \times 2}{9.8 \frac{m}{s^{2}}}

t \approx 9.33 s

Now to find the speed,

v = gt

v\ =\ 9.8\ \frac{m}{s^2}\times9.33\ s\ \approx91\ \frac{m}{x}

The velocity at which it hits the ground is the same as its speed, pointing downwards (direction)

I found an answer from downloads.cs.stanford.edu

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