Question 293547: An object is propelled vertically upward with an initial velocity of 20 meters per second. The distance s (in meters) of the object from the ground after t seconds is s = -4.9t2 + 20t (
a) when will the object be 15meters above the ground?
(b) when will it strike the ground
(c) will the object reach a height of 100 meters?
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! An object is propelled vertically upward with an initial velocity of 20 meters per second. The distance s (in meters) of the object from the ground after t seconds is s = -4.9t2 + 20t
(a) when will the object be 15meters above the ground?
Set s = 15 and solve for t:
15 = -4.9t2 + 20t
0 = -4.9t2 + 20t -15
0 = 4.9t2 - 20t + 15
Solve by applying the quadratic equation which will then give:
t = {3.073, 0.992}
That is, the ball will be at 15 meters
0.992 sec
and again at
3.073 sec
.
(b) when will it strike the ground
Set s = 0 and solve for t:
0 = -4.9t2 + 20t
0 = t(-4.9t + 20)
.
-4.9t + 20 = 0
-4.9t = -20
t = -20/-4.9
t = 4.082 sec
.
(c) will the object reach a height of 100 meters?
Max height is at the "axis of symmetry:
t = -b/2a
t = -20/(2(-4.9))
t = -20/(-9.8)
t = 2.041 sec
.
Height is at:
s = -4.9(2.041)^2 + 20(2.041)
s = 20.41 meters (highest it will go)
.
So, NO, it will not reach 100 meters
.
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