SOLUTION: The height in feet of a projectile with an initial velocity of 160 feet per second and an initial height of 55 feet is a function of time t, in seconds, given by h(t)= -16t^2+160

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Question 323301: The height in feet of a projectile with an initial velocity of 160 feet per second and an initial height of 55 feet is a function of time t, in seconds, given by h(t)= -16t^2+160t+55. Find the maximum height of the projectile.


thanks so much
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
The height in feet of a projectile with an initial velocity of 160 feet per second and an initial height of 55 feet is a function of time t, in seconds, given by h(t)= -16t^2+160t+55. Find the maximum height of the projectile.
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Finding the vertex will give you the max height
h(t)= -16t^2+160t+55
axis of symmetry:
t = -b/(2a)
t = -160/(2(-16))
t = -160/(-32)
t = 5 sec
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So, height is:
h(t)= -16t^2+160t+55
h(5)= -16(5)^2+160(5)+55
h(5) = -16(25)+800+55
h(5) = -400+800+55
h(5) = 400+55
h(5) = 455 feet