Kinematics Ques 43
- 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)
Show Answer
Solution:
Formula:
Projection on an inclined plane:
- 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 $\frac{d t}{d h}=0$
or
$$ \sqrt{\frac{2}{g}} \frac{-1}{2 \sqrt{H-h}}+\frac{1}{2 \sqrt{h}}=0 $$
$$ \begin{aligned} & \text { or } \quad \frac{1}{\sqrt{h}}=\frac{1}{\sqrt{H-h}} \\ & \text { or } \quad 2 h=H \quad \Rightarrow \quad \therefore \quad \frac{h}{H}=\frac{1}{2} \end{aligned} $$
BETA
