An object thrown vertically upwards, rises to a maximum height of $10 m$ . What is the velocity with which it is thrown upwards? (Take $g = 9.8 m s^{- 2}$ )
14

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Solution

The correct option is A 14
Let the upward direction be positive.
Given:
The maximum height reached by the body, $h = 10 m$ .
So, displacement is $s = + h = 10 m$
Acceleration due to gravity acts in downward direction. So,
$a = - g = - 9.8 {m s}^{- 2}$

Let its inital velocity, i.e. the velocity with which it is thrown upwards, be $u$ .
At the highest point, the object changes its direction. So, final velocity, $v = 0$

Using the third equation of motion,
$v^{2} = u^{2} + 2 as$
$0^{2} = u^{2} + 2 \times (- 9.8) \times 10$
$∴ u = \sqrt{196} = 14 m s^{- 1}$

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An object thrown vertically upwards, rises to a maximum height of 10 m. What is the velocity with which it is thrown upwards? (Take g=9.8 m s−2)14

An object thrown vertically upwards, rises to a maximum height of $10 m$ . What is the velocity with which it is thrown upwards? (Take $g = 9.8 m s^{- 2}$ )
14