Question

A particle $A$ is projected vertically upwards. Another particle $B$ is projected at an angle of ${45}^{\circ }$. 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. $\sqrt{2}:1$

Answer 1

Answer: (A)

$1:2$

Let the speed of A and B are $v_1$ and $v_2$, Given, their maximum heights are equal:

$\frac{v_1^2}{2g}=\frac{v_2^{2}si{n}^{2}45}{2g}$

$\Rightarrow v_1^2=\frac{v_2^2}{2}$

$\frac{K_1}{K_2}=\frac{\frac{1}{2}m{v}_1^2}{\frac{1}{2}m{v}_2^2}=\frac{v_1^2}{v_2^2}=\frac{1}{2}$