Astone is thrown straight up from the edge of a roof, 775 feet above the ground, at a speed of 18 feet per second. remembering that the acceleration due to gravity is -32ft/sec^2, how high is the stone 6 seconds later? . at what time does the stone hit the ground? . what is the velocity of the stone when it hits the ground?

We use the formula in1 y = y0 + vo t -0.5at^2 where y0 is the initial height equal to 775 ft, a is -32 ft/s2 and t is equal to 6 sec.substituting, y is equal to 307 feet which means the stone has gone down. in 2, we get the time above 775 which the stone reaches height thru 18  - 32t = 0; t is equal to 0.5625 seconds.  the height above the edge is 780.0625 feet. the time going down is 6.98 seconds. The total time hence is 7.54 seconds. The velocity going down is found through 2ax = vi^2- vo^2. Substituting, 2 (32) *780.0625 = vi2 . Final velocity hence is 223.44 ft/s.

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