Aprojectile launcher upward with a velocity of 160 feet per second from the top of a 70-foot platform. what is the maximum height attained by the projectile?

  470 ft

Step-by-step explanation:

The equation for ballistic motion is usually written ...

  h(t) = -16t^2 +v0·t +h0

where v0 is the initial upward velocity in ft/s, and h0 is the initial height in ft.

Filling in the numbers given in the problem statement, the equation becomes ...

  h(t) = -16t^2 +160t +70

This can be rewritten to vertex form as follows.

  h(t) = -16(t^2 -10t) +70

  = -16(t^2 -10t +25) +70 +400 . . . . . . add and subtract 400 to complete the square

  = -16(t -5)^2 +470

The vertex of the height vs. time curve is at (t, h) = (5, 470).

The maximum height attained is 470 feet.

your answer is the last one

step-by-step explanation:

divide the length of a'c' by the length of ac to get a scale factor of two

& then reflect vertically (over the y-axis)

so your answer is dilation by a scale factor of 2 followed by reflection over the y-axis