Three balls, A, B and C are released and all | vedclass

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Question

(B)

For ball $A$, the vertical velocity is zero, when it leaves the structure. So,

$t_A=\sqrt{\frac{2 h}{g}} \dots(i)$

Also, $\quad t_A=\frac

Vedclass · Accepted Answer

$t_C=t_A=t_C$

Vedclass · Answer

(B)

For ball $A$, the vertical velocity is zero, when it leaves the structure. So,

$t_A=\sqrt{\frac{2 h}{g}} \dots(i)$

Also, $\quad t_A=\frac{x}{v_A}$

where, $v_A=$ horizontal velocity of ball $A$.

From energy conservation,

PE of $A$ at height $h_1= KE$ of $A$

$\Rightarrow m g h_1=\frac{1}{2} m v_A^2$

$\Rightarrow v_A=\sqrt{2 g h_1}$

$	herefore t_A=\frac{x}{\sqrt{2 g h_1}} \dots(ii)$

For ball $B$, using energy conservation,

$m g h_1=\frac{1}{2} m v_B^2$

$\Rightarrow \quad v_B=\sqrt{2 g h_1}$

$	herefore \quad t_B=\frac{x}{v_B}=\frac{x}{\sqrt{2-h_1}} \dots(iii)$

So, $t_A=t_B \quad$ [using Eqs. $(ii)$ and $(iii)$]

For ball $C$, the time can be given for motion under gravity,

$t_C=\sqrt{\frac{2 h}{g}} \dots(iv)$

From Eqs. $(i)$ and $(iv)$, we get

$t_C=t_A=t_B$

Note Correct option can be $(b)$ for $t_A=t_B=t_C$

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