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(C) Let the mass of the bomb be $3m$. At the topmost point,its velocity is $0$. After the explosion,the center of mass continues to follow the origina

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$1250$

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(C) Let the mass of the bomb be $3m$. At the topmost point,its velocity is $0$. After the explosion,the center of mass continues to follow the original trajectory of the bomb,which is a free fall from height $h$ under gravity $g$.Let $v_1, v_2, v_3$ be the velocities of the three fragments. By conservation of momentum,$m v_1 + m v_2 + m v_3 = 0$. Since two fragments take $20 \ s$ to reach the ground,they must have been projected downwards with the same speed $v'$. Let the third fragment be projected upwards with speed $v$. Then $m v - m v' - m v' = 0$,so $v = 2v'$.For the fragment projected upwards: $h = -vt_1 + \frac{1}{2}gt_1^2$ (where $t_1 = 10 \ s$ and $h$ is displacement downwards).For the fragments projected downwards: $h = v't_2 + \frac{1}{2}gt_2^2$ (where $t_2 = 20 \ s$).Substituting $v = 2v'$: $h = -2v'(10) + \frac{1}{2}g(10)^2 = -20v' + 500$.Also $h = v'(20) + \frac{1}{2}g(20)^2 = 20v' + 2000$.Adding the two equations: $2h = 2500 \implies h = 1250 \ m$.

$A$ bomb is projected upwards. At its topmost point,it explodes into three identical fragments. The first fragment reaches the ground in $10 \ s$,and the other two reach the ground in $20 \ s$ each. Find the maximum height $h$ reached by the original bomb. (in $m$)

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