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)   (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?

-------------

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

 

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