Question 1010079: The height h, in meters, at some time, t, in seconds, of an object propelled upward at a speed of 82 meters per second from an initial height of 103 meters is given by the function h(t)=-10t^2+82t+103.
a.) What is the maximum height of the object?
b.) When will the object be above 200 meters high?
c.) When will the object hit the ground?
I got 410 meters for a. I do not understand part b...c I am also unsure.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! -10t^2+82t+103,
The object will be 200 meters high twice, once going up, and once going down.
-10t^2+82t+103=200
10t^2-82t+97=0 moving and multiplying everything by (-1)
t=(1/20)*{82+/- sqrt (6724-3880)}, the sqrt =53.33
t=(1/20)*135.33=6.7665 sec
t=(1/20)*27.67=1.3835 sec
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h=0 when it hits the ground
-10t^2+82t+103=0; 10t^2-82t-103=0
t=(1/20)* {82+/- sqrt (6724+4120); sqrt (10844)=104.13
use positive value only
t=(-1/20)*186.13=9.3 sec.
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maximum height t=-b/2a, and that is -82/-20 or at 4.1 seconds.
h(4.1)=-168.1+336.2+103=271.1 m
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