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

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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

$H_{2}=25 + 10 \sqrt{3}\;.$

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