The water drops fall at regular intervals from a tap $5$ $m$ above the ground. The third drop is leaving the tap at instant the first drop touches the ground. How far above the ground is the second drop at that instant? (Take $g = 10$ $m s^{- 2}$ )

Open in App

Solution

Time taken by $1^{s t}$ drop $= 2 t$
$h = \frac{1}{2} g (2 t)^{2}$
$\Rightarrow t = \sqrt{\frac{h}{2 g}}$
$\Rightarrow t = \frac{1}{2}$ $s$
$h^{'} = \frac{1}{2} g t^{2} = 1.25$ $m$
$∴ h e i g h t = h - h' = 5 - 1.25 = 3.75$ $m$

Suggest Corrections

Similar questions

Q. The water drops fall at regular intervals from a tap

5 m

above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at the instant?

Q. Water drops fall at regular intervals from a tap which is above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant

Water drops fall at regular intervals from a tap which is 5m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant

Q. Water drops fall at regular intervals from a tap 5 m above the ground. How far above the ground is the second drop at that instant when the 3rd drop left the tap and

1_{s t}

drop reaches the ground. (g=

10 m / s^{2}

)

Q. The water drops fall at regular intervals from a tap of 9m above the ground. The 4th drop is leaving the tap at the instant, the first drop touches the ground. How high is the 3rd drop at that instant?

Join BYJU'S Learning Program

The water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at instant the first drop touches the ground. How far above the ground is the second drop at that instant? (Take g=10 ms−2)

Time taken by 1st drop =2th=12g(2t)2⇒t=√h2g⇒t=12 sh′=12gt2=1.25 m∴height=h−h′=5−1.25=3.75 m

The water drops fall at regular intervals from a tap $5$ $m$ above the ground. The third drop is leaving the tap at instant the first drop touches the ground. How far above the ground is the second drop at that instant? (Take $g = 10$ $m s^{- 2}$ )

Time taken by $1^{s t}$ drop $= 2 t$
$h = \frac{1}{2} g (2 t)^{2}$
$\Rightarrow t = \sqrt{\frac{h}{2 g}}$
$\Rightarrow t = \frac{1}{2}$ $s$
$h^{'} = \frac{1}{2} g t^{2} = 1.25$ $m$
$∴ h e i g h t = h - h' = 5 - 1.25 = 3.75$ $m$