Amodel rocket is launched from the ground with an initial velocity of 40 feet per second. the height of an object h, in feet, after t seconds, with initial velocity v0v0 and initial height h0h0 is given by h(t)=−16t2+v0t+h0h(t)=−16t2+v0t+h0 . what is the approximate maximum height the rocket reaches? enter your answer in the box. ft

The rocket reaches a maximum height of 25 feet.

We substitute the velocity given into the equation:
h(t) = -16t² + 40t + h₀

We also use 0 for h₀, as the rocket is launched from the ground:
h(t) = -16t² + 40t + 0

To find the maximum, we find the coordinates of the vertex.  To do this, we first find the axis of symmetry:

-b/2a = -40/(2*-16) = -40/-32 = 1.25

This is the x-coordinate of the vertex, which represents the amount of time in seconds.  We substitute this back into our function:
h(1.25) = -16(1.25²) + 40(1.25) = 25

Shortest side: xside two: x+26side three: 2xwork: x+x+26+2x=1304x+26=130 -26 -264x=104(4x=104)/4x=26check: 26+26+26+52=130shortest side length is 26 feet