In a region an electric field is setup with its strength E =15 N/C and it makes an angle of 30° with the horizontal plane as shown in figure. A ball having a charge 2 C, mass 3 kg and coefficient of restitution with ground 0.5 is projected at an angle of 30° with the horizontal along the direction of electric field with speed 20 m/s. Find the horizontal distance (in m) travelled by ball from first hit with the ground to the second time when it hits the ground. (Use $\sqrt{3}=1.7$)
 

Electric force on ball, $\mathrm{qE}= 30\mathrm{N}$
Vertical component of electric force, ${\mathrm{F}}_{\mathrm{y}}=30 \sin {30}^{\circ }=15\mathrm{N}$
Horizontal component of electric force, ${\mathrm{F}}_x=30 \cos {30}^{\circ }=15\sqrt{3}\mathrm{N}$
$\begin{array}{l}\Rightarrow {\mathrm{a}}_{\mathrm{y}}=\frac{\mathrm{mg}-15}{\mathrm{m}}=\frac{30-15}{3}=5\mathrm{m}/{\mathrm{s}}^2\\ \Rightarrow {\mathrm{a}}_{\mathrm{x}}=\frac{15\sqrt{3}}{3}=5\sqrt{3}\mathrm{m}/{\mathrm{s}}^2\end{array}$
Time of flight for first flight,${\mathrm{T}}_1=\frac{2{\mathrm{u}}_{\mathrm{y}}}{{\mathrm{a}}_{\mathrm{y}}}=\frac{2\times 20 \sin {30}^{\circ }}{5}=4\mathrm{s}$

Time of flight after first hit, ${\mathrm{T}}_2=e{T}_1=0.5\left(4\right)=2\mathrm{s}$
Horizontal velocity after first hit, ${\mathrm{v}}_{{\mathrm{x}}_1}=\left(20\cos {30}^{\circ }\right)+{\mathrm{a}}_{\mathrm{x}}{\mathrm{T}}_1$
$\Rightarrow {\mathrm{v}}_{{\mathrm{x}}_1}=\left(10\sqrt{3}\right)+\left(5\sqrt{3}\right)4=30\sqrt{3}\mathrm{m}/\mathrm{s}$
Horizontal distance travelled between first hit and second hit,
$\begin{array}{l}\mathrm{S}=\left(30\sqrt{3}\right){\mathrm{T}}_2+\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}{\mathrm{T}}_2^2\\ \Rightarrow \mathrm{S}=\left(30\sqrt{3}\right)\left(2\right)+\frac{1}{2}\left(5\sqrt{3}\right)(2{)}^2=70\sqrt{3}\mathrm{m}=119\mathrm{m}\end{array}$