The height y (in feet) of an arrow t seconds after it is shot from a bow can be modeled by the function y=-16t^2+112t + 2

Question

a.) write the function in vertex form

b.) find the maximum height of the arrow

c.) how long does it take the arrow to hit the ground? round your answer to the nearest second

Answer 1

The given function, y = -16t² + 112t + 2, is the position function. In order to determine the maximum height of the arrow, one needs to know that the arrow reaches its maximum height when it stops moving upwards; therefore, one must find the velocity function. Velocity is the rate of change of position; therefore, velocity is the derivative of the position function with respect to time:

1. y(t) = -16t² + 112t + 2
2. y'(t) = -32t + 112

In order to find the amount of time spent by the arrow in flight, the velocity function must be set to zero and the equation must be solved for t:

1. y'(t) = -32t + 112
2. -32t + 112 = 0
3. -32t = -112
4. t = 3.5 s.

Now that one has found the time spent by the arrow in flight, this value may be used in the position function to determine the maximum height achieved by the arrow:

1. y(t) = -16t² + 112t + 2
2. y(3.5) = -16 (3.5)² + 112 (3.5) + 2
3. y(3.5) = -16 (12.25) + 112 (3.5) + 2
4. y(3.5) = -196 + 392 + 2
5. y(3.5) = 198 ft.

Now, to get the above given equation, y(t) = -16t² + 112t + 2, into vertex form... One must complete the square:

1. y = -16t² + 112t + 2
2. y = -16 (t² - 7t - 1/8)
3. y = -16 [t² - 7t + 49/4 - 49/4 - 1/8]
4. y = -16 [ (t - 7t + 49/4) - 49/4 - 1/8 ]
5. y = -16 [ (t - 7/2)² - 99/8 ]
6. y = -16 (t - 7/2)² + 198

Answer 2

just complete the square to get

y+34 = 16(t+ 3/2)^2

vertex is (-3/2, -34)