Question

# A particle is thrown vertically upwards. Its velocity at half of the height is 10m/s, then the maximum height attained by it will be: (g = 10 m/s2).

A

10 m

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B

20 m

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C

15 m

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D

25 m

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Solution

## The correct option is A 10 mStep (1) Given:The velocity at the half of the height, $v=10m/s$The acceleration due to gravity, $g=10m/{s}^{2}$Step (2) Finding the equation of motion when the particle is at the half of the height:Let the maximum height be $h$ meter.By the equation of motion formulas, we know${v}^{2}={u}^{2}–2gs$where $v$ is final velocity, $g$ is the acceleration due to gravity and $s$ is displacement.${10}^{2}={u}^{2}–2g\frac{h}{2}\phantom{\rule{0ex}{0ex}}100={u}^{2}–gh\dots \left(i\right)$Step (3) Finding the equation of motion when the particle is at the maximum height:At the maximum height, $v=0$$0={u}^{2}–2gh\phantom{\rule{0ex}{0ex}}{u}^{2}=2gh\dots ..\left(ii\right)$Step (4) Finding the maximum height using the equations (i) & (ii):Putting the value of u2 in equation (i) we get$100=2gh–gh\phantom{\rule{0ex}{0ex}}100=gh\phantom{\rule{0ex}{0ex}}h=\frac{100}{g}\phantom{\rule{0ex}{0ex}}=\frac{100}{10}\phantom{\rule{0ex}{0ex}}=10m$Hence, the answer is (A) 10 m

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