Question

A cracker moving vertically upward explodes just as it reaches the top of its trajectory at height h above the ground. The display consisting of many identical particles spread out as an expanding brilliant sphere. The bottom of the sphere just touches the ground when its radius is R. Speed of luminous particles ejected by the explosion is given by:
A. $u = R\sqrt {\dfrac{{2g}}{{h - R}}}$
B. $u = h\sqrt {\dfrac{g}{{2\left( {h - R} \right)}}}$
C. $u = R\sqrt {\dfrac{g}{{2\left( {h - R} \right)}}}$
D. $u = h\sqrt {\dfrac{{2g}}{{h - R}}}$

Answer 1

We will calculate the time from the equation of motion in a straight line. Then, the speed of the luminous particles is calculated from the kinematic equation of motion and the particle performs parabolic motion in this time t.

Complete step by step answer:
A cracker initially moves upwards in a vertical direction and it explodes into many identical particles when it reaches at height h. At height h, the luminous particles execute horizontal projectile motion. Let x and y be the horizontal and vertical distance covered by the particles in time t. The horizontal motion of the particles is uniform motion because the only force acting on the particles is force of gravity. This force acts in the vertically downward direction and its horizontal component is zero. So, the equation of motion in horizontal direction is the equation of motion in a straight line i.e., $x = ut$ where u is the speed of the luminous particles ejected by the explosion of the cracker. Hence, $t = \dfrac{x}{u}$.Now, the vertical motion is due to the force of gravity and initial velocity (v) in the downward direction is zero. The acceleration due to gravity (g) is positive as the particles are moving downwards towards the gravity. From kinematic equation,
$y = vt + \dfrac{1}{2}g{t^2}$
$y = 0\left( {\dfrac{x}{u}} \right) + \dfrac{1}{2}g{\left( {\dfrac{x}{u}} \right)^2}\left[ {\because t = \dfrac{x}{u}} \right]$
$h - R = 0 + \dfrac{{g{x^2}}}{{2{u^2}}}$ [ y = h-R because the sphere touches the ground when its radius is R, so the net distance is h-R]
$2{u^2} = \dfrac{{g{R^2}}}{{h - R}}$ [ $x = R$ as horizontal distance is R i.e., radius of the sphere]
$\implies {u^2} = \dfrac{{g{R^2}}}{{2\left( {h - R} \right)}}$
$\implies u = \sqrt {\dfrac{{g{R^2}}}{{2\left( {h - R} \right)}}}$
$\therefore u = R\sqrt {\dfrac{g}{{2\left( {h - R} \right)}}}$.

So, the correct answer is “Option B”.

Note:
When a body is thrown horizontally as when a cracker explodes all the particles are ejected horizontally from a certain height above the ground, they follow a parabolic trajectory till it hits the ground. The initial velocity along downward direction is always zero in this case.