Question

The height of an object that is dropped or thrown straight up or down is given by the quadratic function h(x) = –16x^2+ v0x + h0, where x is the time in seconds, v0 is the velocity at 0 seconds, and h0 is the height at 0
seconds.

Suppose that a ball is thrown straight up from a height of 2 feet at a velocity of 31 feet per second.

Answer 1

What is the question?

Answer 2

what is the maximum height of the ball ?
what time does the ball reach the maximum height ?

Answer 3

ok, h0 is the initial height and v0 is the initial velocity. what does our equation look now when you put that numbers?

Answer 4

h(x)=-16^2+31x+2

Answer 5

i forgot the x in between 16 and 2

Answer 6

yes, this is the equation of a parabola that looks like this:

When that function reaches the maximum?
when h'(x)=0. Can you solve it?

Answer 7

Yes ?

Answer 8

What is derivative of that function?

Answer 9

Is it -2 and 3

Answer 10

are you familiar with calculus?

Answer 11

No

Answer 12

ok, do you know how to find the vertex of parabola? that would be your max height

Answer 13

the vertex is 2 , right ?

Answer 14

no it's \[-\frac{ \Delta}{ 4a }\]

Answer 15

the quadratic formula ?

Answer 16

or substitute x=-b/2a

Answer 17

you should find hmax=31/32

Answer 18

Yes i got x =31/32 or 1089/64

Answer 19

you shouldget only 31/32

Answer 20

yes

Answer 21

then put that in place of h and solve for x. If you find two solutions choose positive one(time cannot be negative)

Answer 22

so im replacing 2

Answer 23

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