Applied Maths!?

Question

a tap drops water bubbles in same time intervals.when drop "A" is dropping from the tap, drop "B" is 0.25m away(towards the earth vertiacally)from the tap.when the distance between drop "A" and "B" becomes 0.75m count the distanse of drop "B" from the tap.
gravity acc.(g)=10ms^-2

Answer 1

drop A is dropping. the drop B is 0.25 m away so its velocity at this point is sqrt 2*10*0.25 =sqrt 5. now when A starts to drop it is also being acted upon by g. So we can assume that B is Moving away from A with a constant velocity sqrt5. The distance between them will become 0.50 more in time t=0.50/sqrt5.
so the distance of drop B from the tap is given by s=ut+1/2gt^2
= sqrt5*0.5/(sqrt5)+0.5*10*(0.50/sqrt5)^2
= 0.5+5*.25/(5)
= 0.5+.25
= 0.75m
now i'm confused.
oh yes you have to add the initial 0.25 because the distance travelled is from the point when the drp was .25m away
so the answer is 1m.

Answer 2

Well, your formula is d = 1/2*gt² = 5t²
When drop B is 0.25m from tap, you have
0.25 = 5t²
0.05 = t²
0.223 = t, which is the time interval between drops.

So drop B's distance is 5t², and drop A's distance is 5(t - 0.223)², and the distance between them is
5t² - 5(t - 0.223)² =
5t² - 5(t² - 0.447t + 0.05) =
5t² - 5t² + 2.236t - 0.25 =
2.236t - 0.25

When this distance is 0.75m,
2.236t - 0.25 = 0.75
2.236t = 1
t = 1/2.236 = 0.447s

and at that time B's distance from the tap will be
d = 5t² = 5(0.447)² = 1m

Did you need to solve it algebraically and exactly? If so, your g value wasn't very exact, was it?

Answer 3

We can always determine the position of a drop be the drop by the ec X = a/2 * t^2
So Dt (dif of t between Drop A an B) at 25m is 0.25 = a/2 * Dt^2
Dt^2= 0.05 Dt= sqrt(.05) = 0.05 ^.5

0.75 = Xb-Xa = a/2 * t^2 - a/2 * (t-Dt)^2 =
a/2 *(t^2 - (t^2 -2t*Dt +Dt^2) = a/2 * (+2t*Dt - Dt^2)
.75 *2/a = (2t * sqrt0.05- 0.05)
t= 0.2/sqrt0.05 /2= aprox 0.447 seg

Xb = a/2 * (0.1/sqrt0.05)^2 = 10/2 *.1^2 /.05 = 1 m

Xa = 1m -.75m = 0.25m

Answer 4

um a waste of water?