A particle is dropped from a height h. A cons | vedclass

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(A) Let the particle be projected horizontally with velocity $u$ at $t = 0$. The horizontal component of velocity remains constant, $v_x = u$. The ver

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(A) Let the particle be projected horizontally with velocity $u$ at $t = 0$. The horizontal component of velocity remains constant, $v_x = u$. The vertical component of velocity at time $t$ is $v_y = gt$. The total velocity $v$ is given by $v^2 = v_x^2 + v_y^2 = u^2 + (gt)^2$. The kinetic energy $E$ is given by $E = \frac{1}{2}mv^2 = \frac{1}{2}m(u^2 + g^2t^2)$. This equation represents a parabola of the form $E = At^2 + B$, where $A = \frac{1}{2}mg^2$ and $B = \frac{1}{2}mu^2$. Since $u > 0$, the initial kinetic energy at $t = 0$ is $E_0 = \frac{1}{2}mu^2 > 0$. As $t$ increases, $E$ increases parabolically. Therefore, the correct graph is shown in option $A$.

$A$ particle is dropped from a height $h$. $A$ constant horizontal velocity $u$ is given to the particle. Taking $g$ to be constant everywhere, the kinetic energy $E$ of the particle with respect to time $t$ is correctly shown in:

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