Question

A particle is uncharged and is thrown vertically upward from ground level with a speed of $5\sqrt{5}{\text{ms}}^{-1}.$ 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 $13{\text{ms}}^{-1}.$ Finally, the particle is given a negative charge $-q.$ Ignoring air resistance. determine the speed (in ${\text{ms}}^{-1}$) 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 $g=10{\text{ms}}^{-2}$).

Answer 1

For uncharged particle $(Q=0)$, net acceleration is only due to gravitation
$u=5\sqrt{5}{\text{ms}}^{-1}$
$h=\frac{u^2}{2g}=\frac{125}{20}\text{m}$
For positive charge $(Q=+q),$ net acceleration be $a_1=a+g,$ where $a$ is acceleration due to electric field
$u=13{\text{ms}}^{-1}$
$h=\frac{125}{20}\text{m}$
$a_1=a+g=\frac{u^2}{2h}=\frac{169\times 20}{2\times 125}=13.52{\text{ms}}^{-2}$
$\Rightarrow a=3.52{\text{ms}}^{-2}$
For negative charge $(Q=-q)$,net acceleration be $a_2=g-a,$ where $a$ is acceleration due to electric field
$h=\frac{125}{20}\text{m}$
$a_2=(10-3.52){\text{ms}}^{-2}=6.48{\text{ms}}^{-2}$
Using $u=\sqrt{2ah}$
$u=\sqrt{\frac{2\times 6.48\times 125}{20}}$
$u=9{\text{ms}}^{-1}$