JEE Advance - Physics (2018 - Paper 2 Offline - No. 9)
A ball is projected from the ground at an angle of $${45^o}$$ with the horizontal surface. It reaches a maximum height of $$120$$ $$m$$ and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of $${30^o}$$ with the horizontal surface. The maximium height it reaches after the bounce, in metres, is ______________.
답변
30
설명
$$H = {{{u^2}{{\sin }^2}(45^\circ )} \over {2g}} = 120$$ 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