Question

Please help, need it so much!
[9.06] Jacob kicks a soccer ball off the ground and in the air with an initial velocity of 33 feet per second. Using the formula H(t) = −16t2 + vt + s, what is the maximum height the soccer ball reaches?

15.1 feet

16.5 feet

17.0 feet

18.2 feet

Answer 1

I don't kno if I am rite, but is it the first one?

Answer 2

well, when you want the maximum or minimum of any function you differentiate and put it equal to zero and find out what that variable was, but i think its a different matter here, give me a sec.

Answer 3

Ok, I will, thx. :)

Answer 4

what is the formula?

Answer 5

H(t) = −16t2 + vt + s

Answer 6

as in what is H(t)? height at given time?

Answer 7

Yes, I think so. :)

Answer 8

It is a height/ time formula. :)

Answer 9

what is the variable s?

is this the full question you dont mind me asking,

Answer 10

Okay, were you given s?

or is s=0?

Answer 11

I think it is seconds. :) Yes, it is the full ?, why?

Answer 12

Wait, when you answer, I will come back later and solve it. :)

Answer 13

s is distance, no?



ok, i will try to answer your question but it will require your judgement, if i did it wrong or not..






H(t)=-16t^2+vt+s


okay, if we differentiate (wrt t) and set to 0 then we can find the maximum, right?



so...


d H(t) =
dt
-32t+v+(d/dt)[s] ---> differentiate s with respect to t


we know that if we differentiate s wrt t we get velocity, right?


sooooo..

the equation becomes -32t+v+v
which is -32t+2v


now we set it to zero

and we know v, which is 33 feet per second,

so,

-32t+66=0

solve for t which is

t=66/32


now this is the time which the ball is at its maximum,


subbing this in the function H(t)...

Answer 14

this is how far i got,
i hope if someone else answers this question maybe they will carry on from there



How did you get your answer to be the first one?

Answer 15

Hmm...I don't know actually, I basically just guessed. :)

Answer 16

I got it,


s is distance, no? s=0 as its being hit from the ground!










H(t)=-16t^2+vt+s


okay, if we differentiate (wrt t) and set to 0 then we can find the maximum, right?



so...


d H(t) =
dt
-32t+v




sooooo..

the equation becomes -32t+v



now we set it to zero

and we know v, which is 33 feet per second,

so,

-32t+33=0

solve for t which is

t=33/32


now this is the time which the ball is at its maximum,


subbing this in the function H(t)...


we get H[33/32]= -16*(33/32)^2 +(33/32)*33 + 0

which is 17.0156,


im still not too sure as it seems sooooooo trivial!!!

Answer 17

Ok, thx so much! This is really hard stuff, lol!