SOLUTION: The height in feet of a projectile with an initial velocity of 96 feet per second and an initial height of 256 feet is a function of time in seconds given by
h(t) = −16t2 +
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Question 1131808: The height in feet of a projectile with an initial velocity of 96 feet per second and an initial height of 256 feet is a function of time in seconds given by
h(t) = −16t2 + 96t + 256.
(a) Find the maximum height of the projectile._______ft
(b) Find the time t when the projectile achieves its maximum height.
t =_____________sec
(c) Find the time t when the projectile has a height of 0 feet.
t =___________sec
Found 2 solutions by Boreal, MathLover1:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
maximum height is vertex and occurs at t=-b/2a, the formula for the vertex of a quadratic.
This is -96/-32 or 3 seconds ANSWER for b.
h(3)=-144+288+256=400 feet ANSWER
c. changing all signs, 16t^2-96t-256=0
divide by 16, t^2-6t-16=0
(t-8)(t+2)=0
t=8 seconds, only positive root ANSWER.
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
(a) Find the maximum height of the projectile._______ft
-> the maximum is at vertex
...complete square
=> and
so, vertex is at (,)
the maximum height of the projectile.___ft
(b) Find the time when the projectile achieves its maximum height.
The value is or seconds
(c) Find the time when the projectile has a height of feet.
......divide by
seconds, (need only positive root )
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