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## Homework Statement

A man standing 31 m from the base of a vertical cliff throws a ball with a speed of 38 m/s aimed directly at a point 5 m above the base of the cliff.

Here is a link to the diagram....

https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-04-2D-Motion/ball_to_cliff/8.gif [Broken]

How long does it take the ball to reach the cliff?

## Homework Equations

First off i tried to find the time to get to the max height:

v = v(initial)+at

0=38+(-9.8)t t=3.87755

Then I found the max height:

x(final) = x(initial) + v(initial)t + 0.5at^2

x(final) = 0 + 38(3.877) + 0.5(-9.8)(3.877^2) x(final) = 73.67

This really does not look right to me though. If it is then would I have to find the time it drops from the max height to 5m? Is there anyway of finding the y component?

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