Question

A shot putter with a mass of `80kg` pushes the iron ball of mass of `6kg` from a standing position accelerating it uniformly form rest at an angle of `45^(@)` with the horizontal during a time interval of 0.1 seconds. The ball leaves his hand when it `2m` high above the level ground and hits the ground 2 seconds later `(g =10m//s^(2))`.
The horizontal disatnce between the point of release and the point where the ball hits the ground .
A. `16 m`
B. `18m`
C. `20m`
D. `22m`

Answer 1

Correct Answer - B
Time of flight `T = 2sec` From `S =ut + (1)/(2) at^(2)`
`-2 = u xx (1)/(sqrt2) xx 2- (1)/(2) xx 10 xx (2)^(2)`
`rArr u =9 sqrt2 m//sec`
The shot putter is gaining momentium to therom ball by applying force (F) for times `(Deltat)`
`FDeltat =mu`
`rArr a Deltat = u rArr a = (u)/(Deltat) = (9sqrt2)/(0.1) =90 sqrt2 m//sec^(2)`
Range `(R) = u cos 45^(@) xx "time of flight"`
`=9 sqrt2 xx (1)/(sqrt2) xx 2 =18m`
The force exerted by shot putter to the ball in horizontal direaction the same force experenced by shot putter in horizontal direction This force is balanced by friction to the shot putter
`F sin 45^(@) =f ----(1)`
`N =mg + f sin 45^(@)---(2)`
`f =mu ---(3)`
From eq (1)(2) and (3)
`rArrmu=(Fcos45^(@))/(mg+Fsin45^(@)),mu(540)/(800+540)`
`:.F =540sqrt2N mu =0.4`

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