SOLUTION: a person standing on the edge of a cliff throws a rock in the air. the path of the rock is described by the function: h(t)= -2t^2 +8t + 6, 2) where h(t) is the height in meters, ab
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-> SOLUTION: a person standing on the edge of a cliff throws a rock in the air. the path of the rock is described by the function: h(t)= -2t^2 +8t + 6, 2) where h(t) is the height in meters, ab
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Question 1181301: a person standing on the edge of a cliff throws a rock in the air. the path of the rock is described by the function: h(t)= -2t^2 +8t + 6, 2) where h(t) is the height in meters, above the ground and t us the time in seconds.
a) what height was the rock thrown into the air
b) determine the vertex of then function by completing the square
c) at what time does the rock reach its optimal height
d) does this this function have a maximum or minimum value? how do you know? state the max/min
e) state the rang of the function as it applies to the real-life model of the situation described. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! h(t)=-2t^2+8t+6.2.
a) This is the constant 6.2 m
b)-2t^2+8t+6.2=h(t)
-2(t^2-4t)+6.2=h(t)
complete the square by taking half of -4 and squaring it, then adding 8 to the constant to balance the equation.
-2(t^2-4t+4)+14.2
-2(t-2)^2+14.2
The vertex is at (2, 14.2), changing the sign of h and keeping the same sign of k.
This is 2 seconds and 14.2 m
c) Not sure what optimal height means in the context of this question but 14.2 m is maximum height
d)Maximum value given the negative quadratic, and that is 14.2 m after 2 seconds
e) The range needs the time it takes to hit or be 0.
so-2t^2+8t+6.2=0
ad t^2-4t-3.1=0, dividing by -2
t=(1/2)(4+/- sqrt (16+12.4)) and sqrt (28.4)=5.33
positive root is 4.66 m. The range of the function is [0, 4.66] units seconds
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Can check this by vertex = -b/2a which is -8/-4 or 2 seconds
h(2)=-8+16+6.2=14.2
Note: the graph is plotting height with time, not distance with time.
