Question

trying to find initial velocity with only initial height and maximum height given, anyone know how?

Answer 1

You could use, if you neglect air resistence, the conservation of mechanical energy.

\[mgh _{0}+1/2mv _{0^{2}}=mgh _{1}\]

Solve for v =)

Answer 2

so, does m represent gravity?

Answer 3

You haven't learned physics, eh? Well, m represents mass, v represents speed and h represents hight. Perhaps you don't know the mass fo the object?

Answer 4

But that doesn't matter, you can divide by m on both sides to get rid of it.

Answer 5

So all you need is initial hight and maximum hight, the rest solves easily.

Answer 6

with g = 9.81m/s^2, gravitational acceleration.

Answer 7

no, just a college algebra paper that's driving me crazy, it's a projectile with a max height or vertex of 400 ft. launched from 30 ft. and we're using -16 for gravitational pull i.e. -16t^2+initial velocity(t)+30

Answer 8

Ah, I am used to working with meters, the whole thing becomes a little different. But the equation I suggested will still look the same, just with g = 16ft/s^2 and the height in feet.

Answer 9

thank you so much, I will give it a go!

Answer 10

If you get into trouble I will outline the solution equation for you ^^

Answer 11

many thanx!

Answer 12

OK, so I'm looking at: \[30+(-16t^{2})=-16t ^{2}*400\] This doesn't make sense to me. Is it simply 30 divided by 400 because the gravity multiplied by time squared just cancels?

Answer 13

I would solve it like this: \[mgh _{0}+(1/2)mgv _{0}^{2} = mgh _{1}\]

Divide by m and multiply by 2:

\[2gh _{0}+v _{0}^{2} = 2gh _{}\]

Move terms and take the square root:

\[v = \sqrt{2gh _{1}-2gh _{0}}\]

Answer 14

that gives me negative 11840, an irrational number, how can this be?

Answer 15

or 108.8117641 squared without the negative symbol

Answer 16

perhaps it is 108 going up and -108 going down?

Answer 17

Ah man, this gets all fluttered up because of different SI units me thinks. What is the equation you have learned to use in your calculus course?

Answer 18

t=time in seconds
-16 ft.=acceleration (downward due to gravity)
s=initial height
v=initial velocity
So, a typical parabola in a quadratic equation might look like: \[f(x)=-16t ^{2}+v(t)+30\] Unfortunately in this case I'm only given -16 for gravity, 30 for initial height and I guess 400 would be F(x) in this scenario.

Answer 19

Well, then just solve for v(t) then. You got all you need

Answer 20

how do I know what to sub in for t?

Answer 21

Haha, true that you, need time. Hmm, this was more delicate than I thought. I will think about it for a while, see if I can help you out better. But the mechanical energy stuff should work, just that you do not put gravity acceleration to be negative. Is the correct answer 108,8 ish?

Answer 22

If I use t=1 seconds, it's v=26.66. If I use t=0 second, it's 13.33. But t cannot be 0 and still have a velocity right? When I used your formula under the radical it was 108ish.

Answer 23

Well, you do want the velocity at time zero, don't you? It's the initial velocity, so plugging in t = 0 could be just fine.

Answer 24

I think I may have just had an epiphany, if 400 represents the point in which the object is neither going up nor down, then v=0. Now I can find t when v=0 and solve! Do you think this will work?

Answer 25

Oh snap, yes ofc it will :)

Answer 26

We made it a lot more difficult than necessary. I am sorry about that, guess I'm not to much help this friday evening^^

Answer 27

No, getting a different look at it really helped and made me think in a different way, thanx again for your help and efforts! I have a tendancy to make things more difficult than they are!

Answer 28

I have a tendency to do that too ;D

Answer 29

v=153.88 ; just so you can sleep tonight :)