Question 1126080
Me:
Ah this should be a problem posted on physics.com...
Checked physics.com
Nope
Better answer it here, though it should be categorized as "Classic Physics"
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Regardless of what's happenning horizontally, vertically its initial speed is 0. It has the accelaration(gravity) of roughly 9.8 m/s^2
The vertical displacement: y=gt^2/2
y=19.6 m(since totally it goes vertically 19.6 metres), g=9.8 m/s^2. Solve. t=2 s.
So the time from start to hitting water is 2 s. This won't change regardless of what's going on horizontally due to the natural properties of these vectors.

Go back. Horizontally it travelled x=vt=2*25=50 metres.

Pretty far!

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Anyway hope that helps! And not to be misleading, feel free to ask any physics qustions here, because they are, fundamentally, applications of math, and algebra.