Water drops fall at regular intervals from a tap which is $5 ext{ m}$ above the ground. The third drop is le

Question

Water drops fall at regular intervals from a tap which is $5 	ext{ m}$ 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 that instant?

Accepted Answer

1. **Analyze Time Intervals:** Let the time interval between consecutive drops be $t$. 
   - The 3rd drop is just leaving, so its time of fall is $0$. 
   - The 2nd drop has been falling for one interval ($t$). 
   - The 1st drop touches the ground, meaning it has fallen for two intervals ($2t$) since the 3rd drop is just starting.
2. **Distance Equation:** Using the equation of motion $s = ut + \frac{1}{2}at^2$ with $u=0$ and $a=g$ .
   - Total height ($H$) fallen by 1st drop in time $2t$: 
     $$ H = \frac{1}{2}g(2t)^2 = \frac{1}{2}g(4t^2) = 2gt^2 = 5 	ext{ m} $$
     From this, $gt^2 = 2.5 	ext{ m}$.
3. **Distance Fallen by 2nd Drop:** The 2nd drop falls for time $t$. Let this distance be $h_2$.
     $$ h_2 = \frac{1}{2}gt^2 $$
     Substituting $gt^2 = 2.5$: 
     $$ h_2 = \frac{1}{2}(2.5) = 1.25 	ext{ m} $$
4. **Calculate Height Above Ground:** The question asks for the height *above the ground*. 
   $$ 	ext{Height} = H - h_2 = 5 	ext{ m} - 1.25 	ext{ m} = 3.75 	ext{ m} $$

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