Question

An arrow is launched upward with a velocity of 192 feet per second from the top of a 95-foot stage. What is the maximum height attained by the arrow?

Answer 1

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Answer:

  671 feet

Step-by-step explanation:

There are a couple of ways to figure this. One is to use a sort of shortcut equation to find the distance traveled (d) by an object when subject to some initial velocity (v) and acceleration (a). Here the acceleration due to gravity is -32 ft/s².

  v² = 2ad

  d = v²/(2a) = (192 ft/s)^2/(2·32 ft/s²) = 576 ft

This height is in addition to the starting height of 95 ft, so the arrow's maximum height is ...

  max height = 95 ft + 576 ft = 671 ft

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Another way to work this problem is to start with the equation for ballistic motion. Filling in the given initial velocity and height, we have ...

  h(t) = -16t^2 +192t +95

The time the arrow reaches the maximum height is the time representing the axis of symmetry of the parabola:

  t = -(192)/(2(-16)) = 6

Then the maximum height is ...

  h(6) = -16·6^2 +192·6 +95 = 671

The maximum height is 671 feet.

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Additional comment

For the standard-form quadratic ...

  y = ax^2 +bx +c

The axis of symmetry is ...

  x = -b/(2a)