QJEEAdv18Phc1b7: JEE Advanced 2018 Physics - Class 11 JEEAdv18Phc1b7 Solution: A ball is projected from the ground at an angle of with the…

$H = {{{u^2}{{\sin }^2}(45^\circ )} \over {2g}} = 120$ m

$\Rightarrow {{{u^2}} \over {4g}} = 120$ m

If speed is v after the first collision, then speed should remain ${1 \over {\sqrt 2 }}$ times, because kinetic energy has reduced to half.

$\Rightarrow v = {u \over {\sqrt 2 }}$

$\Rightarrow {h_{\max }} = {{{v^2}{{\sin }^2}(30^\circ )} \over {2g}}$

$\Rightarrow {h_{\max }} = {{{{\left( {{u \over {\sqrt 2 }}} \right)}^2}{{\sin }^2}(30^\circ )} \over {2g}}$

$\Rightarrow {h_{\max }} = \left( {{{{u^2}/4g} \over 4}} \right) = {{120} \over 4}$

$\Rightarrow {h_{\max }} = 30$ m