Question

I'm not sure how to start this one. Acceleration= 9.8 m/s^2.

Water drips from the nozzle of a shower onto the floor 182 cm below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. When the first drop strikes the floor, how far (in cm) below the nozzle are the (a) second and (b) third drops?

Answer 1

i am a bit confused myself, but i think we can do this:
\[a=-9.8\] ( it should be negative, because gravity sucks)
then
\[v(t)=-9.8t\] by integrating
the constant is zero, because the initial velocity (i assume) is 0
to
\[p(t)=-4.9t^2+182\] and now you can see how long it take for the drop to hit the floor by setting
\[-4.9t^2=182=0\] and solving for \(t\)

sound reasonable?

Answer 2

shouldn't a be positive because it's heading towards the center of the earth?

Answer 3

and that would make the value of t a real number rather than an imaginary number

Answer 4

\[-4.9t^2+182=0\]
\[182=4.9t^2\] etc
it is negative because it is going down

Answer 5

and the solutions are not imaginary

Answer 6

oh oops

Answer 7

The thing is, at my school we are being taught that if for instance, the ball goes down it has a gravity of 9.8 m/s^2

Answer 8

or maybe i just misheard

Answer 9

thanks anyway

Answer 10

I think you could just do this.