SOLUTION: April shoots an arrow upward into the air at a speed of 64 ft per second from a platform that is 34 feet high. The height of the arrow is given by the function h(t)=-16t^2+64t+34,

Algebra ->  Radicals -> SOLUTION: April shoots an arrow upward into the air at a speed of 64 ft per second from a platform that is 34 feet high. The height of the arrow is given by the function h(t)=-16t^2+64t+34,       Log On


   

Question 427100: April shoots an arrow upward into the air at a speed of 64 ft per second from a platform that is 34 feet high. The height of the arrow is given by the function h(t)=-16t^2+64t+34, where t is the time in seconds. What is the maximum height of the arrow?
Found 2 solutions by ewatrrr, Alan3354:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

The height of the arrow is given by the function h(t)=-16t^2+64t+34, 

where t is the time in seconds. What is the maximum height of the arrow?

-16t^2+64t+34    |parbola opening downward, completing the square to find vertex

h(t)=-16t^2+64t+34,

h(t)=-16(t-2)^2 + 64 +34

h(t)=-16(t-2)^2 + 98   |Vertex(2,98)

  98ft is the maximum height of the arrow

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
April shoots an arrow upward into the air at a speed of 64 ft per second from a platform that is 34 feet high. The height of the arrow is given by the function h(t)=-16t^2+64t+34, where t is the time in seconds. What is the maximum height of the arrow?
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The max is the vertex of the parabola, on the line of symmetry t = -b/2a
t = -64/-32 = 2 seconds
h(2) = -16*4 + 64*2 + 34
h(2) = 98 feet, the max ht