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

Algebra ->  Linear-equations -> 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      Log On


   

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) About Me  (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
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In this problem: x=t, a=-4.9, b=4.9, c=7
x+=+%28-4.9+%2B-+sqrt%284.9%5E2-4%2A-4.9%2A7+%29%29%2F%282%2A-4.9%29+
you can do the math here, the positive solution
t = 1.7956 sec
Interval above 2.8 m: 0 to < 1.7956 sec