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April 24, 2014

Homework Help: Physics

Posted by Anonymous on Saturday, March 15, 2008 at 3:21pm.

An object is thrown from the top edge of a building at a speed of 21.21 m/s at an angle of 45degrees above the horizontal. Some time later it passes point A which is 20m below the level of the top of the building, and continues to strike ground at point B. It reaches B moving at an angle of 18.43degrees to the vertical. Determine:

a) the time taken to move to A

b) the distance between A and the building

c) the horizontal and vertical components of velocity at B.

d) the time required to move from A to B

e) the object's speed at B.

So I started out by finding the initial velocity components and the horizontal and vertical components are both 15m/s because 21.21sin45 = 15 and 21.21cos45 = 15.

At A the parallel component remains the same so it will be 15m/s. Not sure how to find the rest of the info as I only know that it's 20m below the level of the top of building...

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