Question

Closed

Answer 1

The time for the water to reach it's peak (where V2 = 0) is
t = v1/a = v1sin(theta)/g
The time for it to travel in the horizontal direction, a distance d is:
t = d/v1cos(theta)

I thought I had it. I'm going to have to think about this a bit longer.

Answer 2

So you subtract the horizontal time from the vertical time, and that's how long it will fall from it's maximum height. Figure out the maximum height in terms of your variables, and you've got it.

Answer 3

max height it reaches would be

\[\frac{(V_i)^{2}\sin^{2}(\theta)}{2(9.8)}\]

Answer 4

according to the formula

Answer 5

I'm still lost blah

Answer 6

That's what I got.
I think I got that backward ... subtract the vertical time from the horizontal time.

Answer 7

how will subtracting max height by d solve this question

Answer 8

I'm talking about the time the water has to fall from the maximum height to the side of the building. Then you can determine how high on the building it will hit by how long it's falling from maximum height.

Answer 9

d = max height + v1t + 1/2at2^
v1 = 0 when it's at max height
So you just need the time it's falling before it hits the building

Answer 10

I'm sorry I'm just too stressed out right now to think, I'm just going to go to sleep I will look over this in the morning hopefully it will make more sense to me then.

Answer 11

One more try:
The total time can be found from the horizontal velocity going a distance d.
Then you subtract the time it takes to reach the maximum height, and you have the time it will fall from maximum height to the height d on the building.

Answer 12

ok so you have

d = 0 + Vi(t) + at^(2)
so you just use the quadratic formula to solve for t

then
hm = 0 + 0(t) + at^(2)
(hm(/a)^(1/2) = t

where hm is just the equation up above.

so then

distance time found using quadratic - (hm/a)^(1/2) = time till it hits wall we will call s

then I can just sub s into

the equation

h = 0 + Vi(s) + a(s)^(2)

Answer 13

wait Vi would be, Visin(Theta) = Vyi

Answer 14

does that seem correct?

Answer 15

My first post I wrote out the two times. It didn't require a quadratic.

Answer 16

wait, I made a mistake shouldn't it be

h = hm + Vi(s) + a(s)^(2)

Answer 17

im out of it I dont think I can figure this out even if you were clear on how to get a solution

Answer 18

Horizontal:
x2 = x1 + vt
d = vcos(theta)t
t = d/vcos(theta)

Answer 19

Vertical:
v2 = v1 + at v2 = 0
t = vsin(theta)/g

Answer 20

I'll sum it up, then it's off to sleep for me also!
The time for it to free fall from maximum height:
t horizontal - t vertical = d/vcos(theta) - vsin(theta)/g

It's a mess, but plug that into the equation for it falling from maximum height to h:
y2 = y1 + v1t + 1/2 at^2
h = max height - 1/2g(d/vcos(theta) - vsin(theta)/g)^2

Answer 21

thanks

Answer 22

Remember that in the horizontal direction, the velocity is constant. There's no acceleration, so the 1/2 at^2 drops out of the equation, and there's no quadratic.
In the vertical direction, you can just use v2 = v1 + at
to find the time, with v2 = 0 at the peak.