SOLUTION: An object launched upward with an initial velocity of 4.9 m/s from a height of 9.8 m will have a height of S(t)=-4.9t^2+4.9t+9.8, where S is in meters and t is in seconds. a)how l

Question 457154: An object launched upward with an initial velocity of 4.9 m/s from a height of 9.8 m will have a height of S(t)=-4.9t^2+4.9t+9.8, where S is in meters and t is in seconds.
a)how long will it take it to hit the ground?
b)Find the interval on which the height of the object is greater than 2.8m.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
An object launched upward with an initial velocity of 4.9 m/s from a height of
9.8 m will have a height of S(t)=-4.9t^2+4.9t+9.8, where S is in meters and t is in seconds.
:
a)how long will it take it to hit the ground?
then S(t) = 0
-4.9t^2 + 4.9t + 9.8 = 0
we can simplify this, divide equation -4.9,
t^2 - t - 2 = 0
Factors easily to
(t-2)(t+1) = 0
positive solution
t = 2 seconds to hit the ground
:
b)Find the interval on which the height of the object is greater than 2.8m.
-4.9t^2 + 4.9t + 9.8 = 2.8
-4.9t^2 + 4.9t + 9.8 - 2.8 = 0
-4.9t^2 + 4.9t + 7 = 0
Use the quadratic formula to find t

In this problem: x=t, a=-4.9, b=4.9, c=7

you can do the math here, the positive solution
t = 1.7956 sec
Interval above 2.8 m: 0 to < 1.7956 sec
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