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?

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Solution

Time taken by the first drop to reach the ground is given by:
$9 m = 9.8 t^{2} \Rightarrow t = 0.96 s$
So, the interval between two drops is 0.32s. So, the distance of third drop from the ground is:
$9 - 9.8 \times {0.32}^{2} = 8 m$

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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}

)

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

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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?

Time taken by the first drop to reach the ground is given by: 9m=9.8t2⇒t=0.96s So, the interval between two drops is 0.32s. So, the distance of third drop from the ground is: 9−9.8×0.322=8m

Time taken by the first drop to reach the ground is given by:
$9 m = 9.8 t^{2} \Rightarrow t = 0.96 s$
So, the interval between two drops is 0.32s. So, the distance of third drop from the ground is:
$9 - 9.8 \times {0.32}^{2} = 8 m$