SOLUTION: Completely stumped on how to do this problem, please help!
Use the position function
s(t) = -16t^2 + v_0t + s_0 **
v_0 = {initial velocity}, s_0 = {initial position}, t
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Question 907280: Completely stumped on how to do this problem, please help!
Use the position function
s(t) = -16t^2 + v_0t + s_0 **
v_0 = {initial velocity}, s_0 = {initial position}, t = {time}
**the v_0 means v sub zero, s_0 means s sub zero, etc.
to answer the following question.
You throw a ball straight up from a rooftop 195 feet high with an initial velocity of 32 feet per second. During which time period will the ball's height exceed that of the rooftop?
ANSWER: Between__ and ___ seconds.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Your initial velocity is given as 32 ft/sec and your initial position (the rooftop) is given as 195 feet. Hence, the position function that describes your situation is:
\ =\ -16t^2\ +\ 32t\ +\ 195)
Your problem is to determine for what range of values of 
is )
more than the height of the roof. In other words \ >\ 195)
, hence you need to solve the quadratic inequality:
Of course, this all assumes that your hand, at the time you released the ball, was exactly at the level of the rooftop. If you are actually more than zero feet tall such that your hand was 
feet above the rooftop when you released the ball, then
is the inequality that you need to solve. In that case, ignore any negative values of because you don't really care what was going on before you threw the ball, do you?
John
My calculator said it, I believe it, that settles it
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