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3 years ago

Please help me out guys. The photo will show you what to do. 50 points cause I need it done fast

Mathematics

1 answer:

Nonamiya [84]3 years ago

6 0

Answer:

a) starting height: 5.5 ft

b) hang time: 5.562 seconds

c) maximum height: 126.5 ft

d) time to maximum height: 2.75 seconds

Step-by-step explanation:

a) The starting height is the height at t=0.

h(0) = -16·0 +88·0 +5.5

h(0) = 5.5

The starting height is 5.5 feet.

b) The ball is in the air between t=0 and the non-zero time when h(t) = 0. We can find the latter by solving ...

-16t^2 + 8t +5.5 = 0

t^2 -(11/2)t = 5.5/16 . . . . . subtract 5.5, then divide by -16

t^2 -(11/2)t +(11/4)^2 = (5.5/16) +(11/4)^2 . . . . complete the square

(t -11/4)^2 = 126.5/16 . . . . . . . . . . . . . . . . . . . . call this [eq1] for later use

t -11/4 = √7.90625

t = 2.75 +√7.90625 ≈ 5.562

The ball will be in the air about 5.562 seconds.

c) If we multiply [eq1] above by -16 and add the constant on the right, we get the vertex form of the height equation:

h(t) = -16(t -11/4) +126.5

The vertex at (2.75, 126.5) tells us ...

The maximum height of the ball is 126.5 feet.

d) That same vertex point tells us ...

The maximum height will be reached at t = 2.75 seconds.

_____

If you really need answers fast, a graphing calculator can give them to you in very short order (less than a minute).

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Given that,

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