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Water drips from a shower head (the sprayer at the top of the shower) and falls onto the floor 2.4 m below. The droplets are falling at regular intervals (equal amounts of time between drops), with the first drop hitting the floor at the instant that the fourth drop starts to fall.When the first drop hits the floor (as the fourth drop is dripping from the shower head), how far below the shower head is the third drop?

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anniwuanyanwu1

Answer:

The third drop is 0.26m

Explanation:

The drop 1 impacts at time T is given by:

T=sqrt(2h/g)

T= sqrt[(2×2.4)/9.8]

T= sqrt(4.8/9.8)

T= sqrt(0.4898)

T= 0.70seconds

4th drops starts at dT=0.70/3= 0.23seconds

The interval between the drops is 0.23seconds

Third drop will fall at t= 0.23

h=1/2gt^2

h= 1/2×9.81×(0.23)^2

h= 0.26m

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