# In a region a horizontal electric field $E = 15N/C$ is present. A ball having charge $2C$ , mass $3kg$ and coefficient of restitution with ground $\;\large\frac{1}{2}\;$ is projected at an angle $\;30^{\circ}\;$ with the horizontal distance travelled by ball from first drop to the second drop

$(a)\;25+5 \sqrt{3}\qquad(b)\;25+10 \sqrt{3} \qquad(c)\;20+5 \sqrt{3} \qquad(d)\;20 +10 \sqrt{3}$

Answer : (b) $\;25+10 \sqrt{3}$
Explanation :
Time taken by particle till $\;1^{st}\;$ drop
$T_{1}=\large\frac{2u sin \theta }{g}=\large\frac{2\times20\times\large\frac{1}{2}}{10}$
$T_{1}=2s$
After hitting the ground its vertical velocity becomes
$v_{f ,{vertical}}=\large\frac{v_{i , vertical}}{2}$
$=\large\frac{20 \times\large\frac{1}{2}}{2}$$=5 m/s Time taken by particle from \;1^{st}\; drop to \;2^{nd}\; drop T_{2}=\large\frac{2\times5}{10}$$=1 s$
Horizontal distance travelled by particle in $\;T_{1}\;$ seconds
$H_{1}=u_{x} T_{1}+\large\frac{1}{2}$$a_{x} T_{1}^{2} u_{x}=20\times\large\frac{\sqrt{3}}{2}$$=10 \sqrt{3} m/s$
$a_{x}=\large\frac{qE}{m}=\large\frac{2\times15}{3}$$=10 m/s^2 H_{1}=10 \sqrt{3}\times2 + \large\frac{1}{2}$$\;\times10\times4$
$H_{1}=20 \sqrt{3}+20$
Horizontal distance by particle in $\;(T_{1}+T_{2})\;$ seconds
$H_{1+2}=u_{x}\;(T_{1}+T_{2})$$+\large\frac{1}{2}$$a_{x} (T_{1}+T_{2})^2$
$=10 \sqrt{3}\times3+\large\frac{1}{2}$$\times 10 \times (3)^2$
$H_{1+2}=30 \sqrt{3} +45$
Horizontal distance travelled by particle in $\;T_{2}\;$ seconds
$H_{2}=H_{1+2}-H_{1}$
$H_{2}=30 \sqrt{3} +45 - (20 \sqrt{3}+20)$
$H_{2}=25 +10 \sqrt{3}$
$H_{2}\;$ is the horizontal distance travelled by particle from $\;1^{st}\;$ drop to $\;2^{nd}\;$ drop & equal to
$H_{2}=25 + 10 \sqrt{3}\;.$
edited Mar 25, 2014