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3-2.Motion in Plane
medium
A particle $A$ is projected vertically upwards. Another identical particle $B$ is projected at an angle of $45^o $ . Both reach the same height. The ratio of the initial kinetic energy of $A$ to that of $B$ is
A
$1:2$
B
$2:1$
C
$1:\sqrt 2$
D
$\sqrt 2:1$
Solution
Let the speed of $A$ and $B$ are $v_{1}$ and $v_{2}$,
Given, their maximum heights are equal$:$
$\frac{v_{1}^{2}}{2 g}=\frac{v_{2}^{2} \sin ^{2} 45}{2 g}$
$\Rightarrow v_{1}^{2}=\frac{v_{2}^{2}}{2}$
$\frac{K_{1}}{K_{2}}=\frac{\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v_{2}^{2}}=\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{2}$
Standard 11
Physics
