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Water drops fall at regular intervals fr...

Water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap, the instant the first drop touches the ground. How far above the ground is the second drop at that instant. `(g = 10 ms^-2)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C

`x=a cospt, y=b sin pt, vec(r)=a cos (pt) hat(i)+b sin (pt) hat(j)`
`:' sin^(2)pt+cos^(2)pt=1`
`:. x^(2)/a^(2)+y^(2)/b^(2)=1` (ellipse)
`vec(v)=-ap sin (pt)hat(i)+bp cos (pt)hat(j), v_(t)=pi/(2p)=-aphat(i)`
`vec(a)=-ap^(2)(pt)hat(i)=bp^(2) sin (pt)hat(j), a_(t)=pi/(2p)=-bp^(2)hat(j)`
`vec(a).vec(v)=0`
`vec(a)=-p^(2) [a cos pthat(i)+b sin pthat(j)]=-p^(2)vec(r)`
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Knowledge Check

  • Water drops are falling from a pipe at 5 m height at regular interval of time. When the third drop is released at the same time the first drop touches the ground. Then the height of second drop from ground is ....... m. (g = 10 ms^(-2) )

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