Kinematics Water Tap Problem Shortcut

  

❓ Question

A tap is at a height of $5m$ above the ground. Water drops fall from it at equal time intervals.

When the 1st drop hits the ground, the 6th drop is just about to fall.

Find the height of the 4th drop above the ground at that instant.

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✍️ Short Explanation

Let the equal time interval between successive drops be $\tau$.

When the 1st drop hits the ground, the 6th drop is just released.

Hence the time taken by the 1st drop to reach the ground is

$$5\tau \mathrm{.}$$

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🔷 Step 1 — Time of Fall of First Drop

For the first drop:

$$5=\frac{1}{2}g(5\tau {)}^2\mathrm{.}$$

We do not need to find $\tau$ explicitly.

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🔷 Step 2 — Time for Which the 4th Drop Has Fallen

The 4th drop is released after $3\tau$ from the first drop.

Therefore, when the first drop hits the ground,

$$t_4=5\tau -3\tau =2\tau \mathrm{.}$$

So the 4th drop has been falling for $2\tau$.

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🔷 Step 3 — Distance Fallen by the 4th Drop

Using

$$s=\frac{1}{2}g{t}^2,$$
$$s_4=\frac{1}{2}g(2\tau {)}^2\mathrm{.}$$

From the first-drop equation,

$$5=\frac{1}{2}g(5\tau {)}^2\mathrm{.}$$

Therefore,

$$\frac{s_4}{5}=\frac{(2\tau {)}^2}{(5\tau {)}^2}=\frac{4}{25}\mathrm{.}$$
$$s_4=5\times \frac{4}{25}=0.8m\mathrm{.}$$

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🔷 Step 4 — Height Above Ground

Initial height = $5m$

$$h_4=5-0.8$$
$$h_4=4.2m\mathrm{.}$$

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✅ Final Answer

$$\boxed{{\displaystyle 4.2\text{ m}}}$$

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🔷 JEE Shortcut

If the 1st drop takes $n$ equal intervals to reach the ground, then a drop that has been falling for $k$ intervals falls:

$$s=H{\left(\frac{k}{n}\right)}^2\mathrm{.}$$

Here,

$$H=5m,n=5,k=2.$$
$$s=5{\left(\frac{2}{5}\right)}^2=0.8m$$
$$\boxed{{\displaystyle \text{Height above ground}=5-0.8=4.2m\mathrm{.}}}$$

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