Posted by Bretagne on Sunday, January 20, 2008 at 2:39pm.
Suppose you are tossing an apple up to a friend on a third story balcony. After t seconds, the height of the apple in feet is given by h = -16t^2 + 38.4t + .96. Your friend catches the apple just as it reaches its highest point. How long does the apple take to reach your friend, and at what height above the ground does your friend catch the apple?
From the height formula, the initial upward speed is 38.4 ft./sec.
From Vf = Vo - gt, Vf = 0, Vo = 38.4 ft./sec. and g = 32 ft/sec.^2.
0 = 38.4 -32t making t = 1.2 sec.
Therefore, h = 38.4(1.2) - 16(1.2)^2 + .96 = 24 ft.
Time to friend is 1.2 sec. at h = 24 ft.
How did you get 32ft for g?
how did you get g equaled to 32ft/sec^2? I'm a little confused on that part?
The earth's surface acceleration due to gravity is g = 32.2 ft/sec^2, rounded off to 32.
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