Water drips from the nozzle of a shower onto the floor 181 cm below. The drops f
ID: 1886312 • Letter: W
Question
Water drips from the nozzle of a shower onto the floor 181 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 below the nozzle are the (a) second and (b) third drops? Water drips from the nozzle of a shower onto the floor 181 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 below the nozzle are the (a) second and (b) third drops?Explanation / Answer
height of the nozzle above floor = 181 cm = 1.81 m
the time taken by the first drop to fall through this distance can be calculated
by using the formula S = u*t + (1/2)*a*t2
1.81 = 0 + 0.5* 9.8 * t 2
t2 = 1.81 / 4.9
t2 = 0.369 s2
t = 0.607 s
when the first drop reaches the floor , the fourth one is about to fall,
the second and third drops are in air.
According to this problem, the drops are falling on the floor at regular intervals of time
that is the time difference between 1st and 2nd = time difference between 2nd and 3rd = time difference between 3rd and 4th
let this interval be t0 , then t0 = (1/3 )*t
= (1/3) * 0.607 s
= 0.202 s
when the first drop has struck the floor , the 2nd drop has fallen for a time 0.404 s , the 3rd one has fallen for a time 0.202 s, while the forth one is just to fall.
(A)
Heigth through which 2nd drop fallen S2 = (1/2 )*g*t12
here, t1 = 2t0 , then S2 = 0.5 * 9.8 * (0.404 s)2
S2 = 0.7997 m
S2 = 79.97 cm
(B)
Heigth through which 3 rd drop fallen S3 = (1/ 2)*g *t22 , here t 2 = t o
S3 = 0.5 * 9.8 * (0.202 s)2
S3 = 0.1999 m
S3 = 19.99 cm
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