In a region, an electric field $E=15N/C$ making an angle of $30°$ with the horizontal plane is present. A ball having charge $2C$, mass $3kg$ and coefficient of restitution with ground $1/2$ is projected at an angle of $30°$ with the horizontal in the direction of electric field with speed $20m/s$. Find the horizontal distance travelled by ball from first drop to the second drop.

$qE=30N$, vertical component of electric force $=30·\sin 30°=15N$ and horizontal-component  electric force $=30\cos 30°=15\sqrt{3}N$

$$\begin{gathered} a_y=\frac{mg-15}{m}=\frac{30-15}{3}=5m/s^2 ( vertically downward) \\ {a}_x=\frac{15\sqrt{3}}{3}=5\sqrt{3}m/s^2 \\ {T}_1=\frac{2u_y}{a_y}=\frac{2\times 20\sin 30°}{5}=4s \\ {T}_2=e{T}_1=2s \end{gathered}$$

Horizontal range after first drop

$$\begin{gathered} =\left(20\cos 30°\right)T_2+\frac{1}{2}{a}_x{T_2}^2 \\ =\left(10\sqrt{3}\right)4+\frac{1}{2}\left(5\sqrt{3}\right)4 \\ =50\sqrt{3}m/s \end{gathered}$$

$\therefore$ Horizontal distance travelled between first drop and second drop

$$\begin{gathered} =\left(10\sqrt{3}\right)T_2+\frac{1}{2}{a}_x{T}_2^2 \\ =\left(10\sqrt{3}\right)\left(2\right)+\frac{1}{2}\left(5\sqrt{3}\right)(2{)}^2 \\ =30\sqrt{3}m \end{gathered}$$