Free Fall - down) are now easy to write. If we locate the...

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Free Fall The first application we will discuss is that of objects in free fall. In general, the acceleration of an object in the earth's gravitational field is not constant. If the object is far away, it will experience a weaker gravitational force than if it is close by. Near the surface of the earth, however, the acceleration due to gravity is approximately constant--and is the same value regardless of the mass of the object (i.e., in the absence of friction from wind resistance, a feather and a grand piano fall at exactly the same rate). This is why we can use our equations for constant acceleration to describe objects in free fall near the earth's surface. The value of this acceleration is a = 9.8 m/s 2 . From now on, however, we will denote this value by g , where g is understood to be the constant 9.8 m/s 2 . (Notice that this is not valid at large distances from the surface of the earth: the moon, for instance, does not accelerate towards us at 9.8 m/s 2 .) The equations describing an object moving perpendicular to the surface of the earth (i.e. up and
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Unformatted text preview: down) are now easy to write. If we locate the origin of our coordinates right at the earth's surface, and denote the positive direction as that which points upwards, we find that: x ( t ) = - gt 2 + v t + x Notice the - sign that arises because the acceleration due to gravity points downwards, while the positive position-direction was chosen to be up. How does this relate to an object in free fall? Well, if you stand at the top of a tower with height h and let go of an object, the initial velocity of the object is v = 0 , while the initial position is x = h . Plugging these values into the above equation we find that the motion of an object falling freely from a height h is given by: x ( t ) = - gt 2 + h If we want to know, for instance, how long it takes for the object to reach the ground, we simply set x ( t ) = 0 and solve for t . We find that at t = the object hits the ground (i.e. reaches the position 0 )....
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This note was uploaded on 02/09/2012 for the course PHY PHY2053 taught by Professor Davidjudd during the Fall '10 term at Broward College.

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