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(A) Let the initial speeds of particles $A$ and $B$ be $v_{1}$ and $v_{2}$ respectively.The maximum height $H$ reached by a particle projected vertica

Vedclass · Accepted Answer

$1:2$

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(A) Let the initial speeds of particles $A$ and $B$ be $v_{1}$ and $v_{2}$ respectively.The maximum height $H$ reached by a particle projected vertically is $H = \frac{v^2}{2g}$.The maximum height $H$ reached by a particle projected at an angle $	heta$ is $H = \frac{v^2 \sin^2 	heta}{2g}$.Given that both reach the same height,we have:$\frac{v_{1}^{2}}{2g} = \frac{v_{2}^{2} \sin^{2} 45^{\circ}}{2g}$Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,we have $\sin^2 45^{\circ} = \frac{1}{2}$.Substituting this into the equation:$v_{1}^{2} = v_{2}^{2} 	imes \frac{1}{2} \Rightarrow \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{2}$.The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$. The ratio of the initial kinetic energy of $A$ to that of $B$ is:$\frac{K_{1}}{K_{2}} = \frac{\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v_{2}^{2}} = \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{2}$.Thus,the ratio is $1:2$.

$A$ particle $A$ is projected vertically upwards. Another identical particle $B$ is projected at an angle of $45^{\circ}$. Both reach the same maximum height. The ratio of the initial kinetic energy of $A$ to that of $B$ is

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