SATHEE: Kinematics 5 Question 19

Kinematics 5 Question 19

18. A body falling freely from a given height $H$ hits an inclined plane in its path at a height $h$ . As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of $(h / H)$ , the body will take maximum time to reach the ground?

(1986, 6M)

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Solution:

Let $t_{1}$ be the time from $A$ to $B$ and $t_{2}$ the time from $B$ to $C$ . Then, $t_{1} = \sqrt{\frac{2 (H - h)}{g}}$ and $t_{2} = \sqrt{\frac{2 h}{g}}$

Then, the total time

$t = t_{1} + t_{2} = \sqrt{\frac{2}{g}} [\sqrt{H - h} + \sqrt{h}]$

For $t$ to be maximum $\frac{d t}{d h} = 0$

$\sqrt{\frac{2}{g}} \frac{- 1}{2 \sqrt{H - h}} + \frac{1}{2 \sqrt{h}} = 0$

$\begin{aligned} or \frac{1}{\sqrt{h}} = \frac{1}{\sqrt{H - h}} \\ or 2 h = H \Rightarrow ∴ \frac{h}{H} = \frac{1}{2} \end{aligned}$

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Kinematics 5 Question 19

18. A body falling freely from a given height H hits an inclined plane in its path at a height h. As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of (h/H), the body will take maximum time to reach the ground?

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18. A body falling freely from a given height $H$ hits an inclined plane in its path at a height $h$ . As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of $(h / H)$ , the body will take maximum time to reach the ground?