Kinematics Ques 12
- From a tower of height $H$, a particle is thrown vertically upwards with a speed $u$. The time taken by the particle to hit the ground, is $n$ times that taken by it to reach the highest point of its path. The relation between $H, u$ and $n$ is
(a) $2 g H=n^{2} u^{2}$
(b) $g H=(n-2)^{2} u^{2}$
(c) $2 g H=n u^{2}(n-2)$
(d) $g H=(n-2)^{2} u^{2}$
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Answer:
Correct Answer: 12.(c)
Solution:
Formula:
Equations of Motion (for constant acceleration):
- Time taken to reach the maximum height, $t _1=\frac{u}{g}$
If $t _2$ is the time taken to hit the ground, then
$$ \begin{array}{rlrl} \text { i.e. } & & -H & =u t _2-\frac{1}{2} g t _2^{2} \\ \text { But } & & t _2=n u _1 \\ \text { So, } & -H & =u \frac{n u}{g}-\frac{1}{2} g \frac{n^{2} u^{2}}{g^{2}} \\ & -H & =\frac{n u^{2}}{g}-\frac{1}{2} \frac{n^{2} u^{2}}{g} \\ \Rightarrow & 2 g H & =n u^{2}(n-2) \end{array} $$
[Given]
BETA
