<PRE><B>Question 763757</B>
f an object is thrown vertically upward with an initial velocity of v, from an original position of s, the height h at any time t is given by: h=-16t^2+vt+s
(where h and s are in ft, t is in seconds and v is in ft/sec) 
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note that an object thrown upward would have a positive value for the initial velocity (v)....in our case the object is thrown downward so the value for v is negative.
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a) what is height after 2 seconds?
h = -16*t^2 +v*t + s
h = -16*(2^2) -5*2 + 300
h = -64 -10 +300
h = 226 feet after 2 seconds
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b) how long does it take to hit the ground?  that is h=0
0 = -16t^2 -5t + 300
divide by -1 and we get
0 = 16t^2 +5 - 300
use quadratic formula
t = (-5+sqrt(25-4*16*-300)) / 32 = 4.18
t = (-5-sqrt(25-4*16*-300)) / 32 = -4.49
t reaches the ground in 4.18 seconds
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