Question

(Maybe I should change my profile to an attractive girl so I can get this answered =P)
You toss a rock up vertically at an initial speed of 30 feet per second and release it at an initial height of 6 feet. The rock will remain in the air for (answer)seconds.
It will reach a maximum height of (answer) feet after (answer) seconds.
Note: Ignore air resistance.

Answer 1

this sounds like physics

Answer 2

It's for a Calculus class.

Answer 3

what formula or equation are you using for this one?

Answer 4

I don't know, since he hasn't really gone over an equation for this, which is why I'm wondering it's even on my homework.

I know that I have to take gravity in to account, which I believe in ft/s is generally 32. Since the rock is already at 6 ft, I know that will be factored in as well.

At one point the rock will stop, and gravity will send it back down, so the 2nd velocity is going to be 32ft/s, and the initial velocity is 30 ft/s.

Answer 5

g= 32 ft/s^2

Answer 6

you can use the equations for constant acceleration which are derived using calculus

Answer 7

we could solve for v0 - gt = 0 to find the time to reach 0 velocity

Answer 8

so, 30 - 32(6)^2?

Answer 9

wait, ignore that 6.

Answer 10

t = -30/32?

Answer 11

30/32*

Answer 12

yes, it takes that long to reach the peak.

distance will be
d= x0 + v0 * t - g t^2
that will be the max height.

Answer 13

so when it reaches the peak, I just double that number?

Answer 14

Where will the 6ft come in to play? x0?

Answer 15

if we left from height 0. But I would be more careful here and work the equations

Answer 16

ya, I just got 1.875 and it took it as wrong, so I know I'm not using that 6 right.

Answer 17

once we find the max height, we can figure out how long it takes to fall that distance.
that will give the time down.

Answer 18

So I have to find the max height first, and then answer the first question last?

Answer 19

yes, at least that is what I'm doing... following my nose on this one.

Answer 20

d= x0 + v0 * t - g t^2

x0 is the start distance = 6
v0 = 30
g = 32

t= 15/16

Answer 21

All right, let me calculate for a second.

Answer 22

After calculating, I got d=6+28.125-28.125, which would be 6, which was wrong.

Answer 23

sorry. what we should do is
\[ \int v - g \ t\ dt = v t - \frac{1}{2} g t^2 + c\]

in other words we should use
d= x0 + v0 * t - (1/2) g t^2

Answer 24

Recalculating...

Answer 25

20.0625, it was correct!

Answer 26

so using that same equation, we find the t?

Answer 27

I would just pretend we dropped it from a height of 20.0625
and use
d= 1/2 g t^2
and solve for t

Answer 28

time up is 0.9375
add time down for total time.

Answer 29

I got 2.0572795542, which it was saying is wrong...

Answer 30

do they ask for you to round ?

Answer 31

doesn't say so, I tried rounded versions.

Answer 32

we can write this equation
h(t) = -1/2 g t^2 + 30 t + 6
and solve for when the height is 0, i.e. solve
-16 t^2 + 30 t + 6 =0
for t

Answer 33

Recalculating...

Answer 34

I got this 1/16 (15+sqrt(321)), for t, and it's a no go on that.

Answer 35

oh wait a minute, I'm an idiot, hold on.

Answer 36

oh wait no, I use quadratic and got that answer, what did I do wrong?

Answer 37

\[ \frac{1}{2a} ( -b \pm \sqrt{b^2 -4ac}) \]

Answer 38

ya, used the quadratic formula.

Answer 39

but it gives the same t as what you found
the good news is
h(t) = -16 t^2 + 30 t + 6

has a peak at t= 30/32= 15/16 (using vertex x value is at -b/(2a) )
and h( 15/16) = 20.065

the only problem seems to be the time at which it hits the ground
2.057...

Answer 40

I'm completely stumped, I'm looking through class notes and don't see anything like this. Great news too, I have another problem just like this after this question for my homework! Looks like I'll be busy for a while.

Answer 41

Do I need to do something with derivates since that's what section we are in class?

Answer 42

we can start with Newton's F= ma and derive the equations we are using
but we will get the same answer.

what *exactly* have you tried typing in for the total time in the air?
they may want just 2
or 2.06
or 2.0573
?

Answer 43

2.0573, 2.05, 2.1, 2.06, and then the WHOLE thing.

Answer 44

2.1 ?

Answer 45

ya...rounded up from 2.06?

Answer 46

It was just a shot in the dark.

Answer 47

or we have a typo and they want the time to reach the peak = 15/16

Answer 48

I have no idea, emailed my professor, hopefully I can get this sorted out. I'll answer back if I get anything. Thanks for the help!

Answer 49

Emailed him and he says I'm not taking the initial 6 feet into account when I do d= 1/2gt^2

Answer 50

if we found the correct max height then we did that part correctly.
maybe we catch the rock? so total time is 1.875

Answer 51

Total time doesn't equal 1.875 according to this, I have 1.875 as total time in the air, and it's giving me a wrong answer.

Answer 52

this is also probably wrong, but the only other "total time" would be if we doubled the down time, to get 2.2396

but the total time from leaving the hand 6 ft off the ground at 30 ft/sec until it hits the ground is 2.06 sec

anything else is a "bug" or typo. Though if your professor can explain it, I'll retract my assertion.

Answer 53

Where did you get 2.06?

Answer 54

the solution to
-16 t^2 + 30 t + 6 =0

Answer 55

What were you using for t?

Answer 56

***
Emailed him and he says I'm not taking the initial 6 feet into account when I do d= 1/2gt^2
****

using that formula would not be correct (though we used it to find the time to drop from the distance of 20.00)

we are using
-16 t^2 + 30 t + 6 =0

which is the equation of height (or distance) above the ground as a function of time.
at time t=0, the height is 6
at time = 15/16 , the height is 20.0625 ft
at time = 2.057, the height is 0 ft (i.e. on the ground)

if we caught the rock 6 ft above the ground we would solve
-16 t^2 + 30 t + 6 = 6
t (-16 t + 30) = 0 --> t=0 or t= 30/16= 1.875
as *possible* value for total time in air. (which you said did not check out)

Answer 57

Here is graph of the problem

Answer 58

.938 is correct! How did you get that?

Answer 59

So, how do I get the answer to the first problem with this information then?

Answer 60

the time from leaving the hand at t=0 until it reaches its max height is t= 15/16 sec
15/16= 0.9375
rounded, 0.938

Answer 61

in other words, 0.9375 sec is the time *on the way up*

NOT the total time in the air. So we have a typo in the question (always makes it hard to answer when we don't know the question)

Answer 62

So, all the answers: total time in the air is 2.057, the point that it reaches 20.0625 feet is .938.

To get the maximum, we used d= x0 + v0 * t - (1/2) g t^2
x0=6, v0=30, g=32, t=30/32

With that, we got the total time in the air by solving for t of d= 1/2 g t^2,
d=20.0625, g = 32.

To get at what time it would reach maximum heigh, we divided initial velocity by gravity.

Answer 63

looking at the top of the post, it is the first number we found... unfortunately, we thought they wanted the total time in the air, not the time to reach the peak height.

Answer 64

they wanted both of those, time it took to reach peak height AND the total air time.

Answer 65

when you posted
.938 is correct! How did you get that?

what question was that for?

Answer 66

For when it would reach the peak height.

Answer 67

After how many seconds.

Answer 68

ok, and did you ever get a correct response for the total time?

Answer 69

Yeah, it was 2.057, looks like we had the things reversed.

Answer 70

we?!

Answer 71

The next question on the homework now is a general problem about ALL of this.
You toss a rock up vertically at an initial speed of V feet per second and release it at an initial height of H feet. The rock will remain in the air for (ANSWER#1) seconds.
It will reach a maximum height of (ANSWER#2) feet after (ANSWER#3) seconds.

Answer 72

hahah, nevermind, I!! I had it reversed, okay?! hahah. Thanks for all your help on that question.

Answer 73

I already got one answer for the next question. Answer # 3 is V/g

Answer 74

for part II, tweak
h(t) = -1/2 g t^2 + 30 t + 6

replace 30 with V (initial velocity) and replace 6 with H (initial height)
-1/2 g = -1/2 * 32 = -16
so

\[ h(t) = -16 t^2 +V t + H \]
you can find the peak by find the derivative and setting equal to 0, and then solve for t
max height will occur at that t

solve for total time by using the quadratic formula on
\[ -16 t^2 +V t + H = 0 \]
i.e. find the time when height is 0 (rock is on the ground)

Answer 75

It's being real knitpicky. Won't let me use "t". I'm only allowed to use V, H, and g in my answers =/

Answer 76

which question are you answering now?

Answer 77

Trying to answer all except the last one. I already got that.