An Open Challenge

Time taken to catch the ball on ground :

$t_1=\dfrac{2u}{g}=\dfrac{(2)(50)}{10}=10s$

Time taken to catch the ball on the block

$t_2=t_1-8=2s$

Let the acceleration of the block be $a$ (wrt the person)

$t_2=\dfrac{2u}{a}=2$

$a=50=5g$

Therefore ,effective acceleration felt by the person is $5g$

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Generalisation for effective acceleration (wrt person) to be 'n' times g

Let the block $A$ move upwards with $(n-1)g$ acceleration

Let the normal force between the wedge and the block be $N$

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FBD of the block A

$N \cos \theta-mg=m(n-1)g$

$N\cos \theta =m(n-1+1)g=m(ng)$ - - - - - - - 1

(ng) $\rightarrow$ shows that acceleration felt by the person is n times g

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FBD of the wedge B

Let the acceleration of wedge be A

$F-N\sin \theta =MA$ - - - - - - - 2

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Condition for blocks to be in contact

For the blocks to be in contact, their normal acceleration should be same

$A\sin \theta = (n-1)g\cos \theta$ - - - - - - - 3

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Equation 2 can be written as

$N\sin \theta=F-MA$ - - - - - - - 4

Divide eq 4 by 1

$\tan \theta=\dfrac{F-MA}{mng}$

$F=mng\tan \theta +MA$

$A=(n-1)g\cot \theta$ (from eq 3 )

$F=mng\tan \theta+M(n-1)g\cot \theta$ - - - - - - - 5

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For minimum F

$\dfrac{dF}{d \theta} = mng\sec ^2 \theta-M(n-1)g {cosec} ^2 \theta=0$

$\dfrac{mng}{\cos ^2 \theta}=\dfrac{M(n-1)g}{\sin ^2 \theta}$

$\dfrac{\sin ^2 \theta}{\cos ^2 \theta}=\dfrac{M(n-1)}{m(n)}$

$\tan \theta=\sqrt{ \dfrac{M(n-1)}{m(n)}} \Rightarrow \cot \theta=\sqrt{ \dfrac{m(n)}{M(n-1)}}$

Substituting $\tan \theta, \cot \theta$ values in eq 5

$F=2g \sqrt{Mmn(n-1)}$

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Now $n=5, m=196kg, M=20kg$

$F=2(10)\sqrt{(20)(5)(196)(4)}=5600N$

$\displaystyle \color{#3D99F6}{ \boxed{ \therefore F=5600N}}$

Shouldn't we be considering the impulse received by the man and and the weight in reaction to throwing up the ball too??