(A) For a body thrown horizontally from a height $H$ with velocity $v$,the horizontal range $R$ is given by $R = v \sqrt{\frac{2H}{g}}$.Given for the

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(A) For a body thrown horizontally from a height $H$ with velocity $v$,the horizontal range $R$ is given by $R = v \sqrt{\frac{2H}{g}}$.Given for the first case: $100 = v \sqrt{\frac{2H}{g}}$.For the second case,the mass is $2M$,the velocity is $v' = \frac{v}{2}$,and the height is $H' = 4H$.The new horizontal range $x$ is given by:$x = v' \sqrt{\frac{2H'}{g}} = \left( \frac{v}{2} \right) \sqrt{\frac{2(4H)}{g}}$$x = \frac{v}{2} \cdot 2 \sqrt{\frac{2H}{g}} = v \sqrt{\frac{2H}{g}}$Since $v \sqrt{\frac{2H}{g}} = 100 \ m$,we get $x = 100 \ m$.

Class 11 Physics 3-2.Motion in Plane Horizontal Projectile Motion

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$A$ body of mass $M$ thrown horizontally with velocity $v$ from the top of a tower of height $H$ touches the ground at a distance of $100 \ m$ from the foot of the tower. $A$ body of mass $2M$ thrown at a velocity $\frac{v}{2}$ from the top of a tower of height $4H$ will touch the ground at a distance of ........ $m$.

JEE MAIN 2024 All JEE MAIN

A
$100$
B
$199$
C
$198$
D
$197$

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3-2.Motion in Plane

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