A particle is uncharged and is thrown vertically upward from ground level with a speed of $5 \sqrt{5} {ms}^{- 1} .$ As a result, it attains a maximum height $h .$ The partide is then given a positive charge $+ q$ and reaches the same maximum height $h$ when thrown vertically upward with a speed of $13 {ms}^{- 1} .$ Finally, the particle is given a negative charge $- q .$ Ignoring air resistance. determine the speed (in ${ms}^{- 1}$ ) with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the same maximum height $h$ . (Assume uniform electric field and $g = 10 {ms}^{- 2}$ ).

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Solution

For uncharged particle $(Q = 0)$ , net acceleration is only due to gravitation
$u = 5 \sqrt{5} {ms}^{- 1}$
$h = \frac{u^{2}}{2 g} = \frac{125}{20} m$
For positive charge $(Q = + q),$ net acceleration be $a_{1} = a + g,$ where $a$ is acceleration due to electric field
$u = 13 {ms}^{- 1}$
$h = \frac{125}{20} m$
$a_{1} = a + g = \frac{u^{2}}{2 h} = \frac{169 \times 20}{2 \times 125} = 13.52 {ms}^{- 2}$
$\Rightarrow a = 3.52 {ms}^{- 2}$
For negative charge $(Q = - q)$ ,net acceleration be $a_{2} = g - a,$ where $a$ is acceleration due to electric field
$h = \frac{125}{20} m$
$a_{2} = (10 - 3.52) {ms}^{- 2} = 6.48 {ms}^{- 2}$
Using $u = \sqrt{2 a h}$
$u = \sqrt{\frac{2 \times 6.48 \times 125}{20}}$
$u = 9 {ms}^{- 1}$

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