Sewag dont hit too hard coz the meter is damaged!

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rakeen
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Sewag dont hit too hard coz the meter is damaged!

Unread post by rakeen » Sat Mar 19, 2011 5:49 pm

Sewag hit a ball(suppose the ball has no velocity) and it goes 102 m from the crease.(suppose there is no wind). What is the initial velocity of the ball and how hard he have to hit it? when $g = 9.8$ and the mass of the ball is 200 g . :roll: :?:
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Re: Sewag dont hit too hard coz the meter is damaged!

Unread post by Moon » Sun Mar 20, 2011 8:09 pm

I didn't really get your question. Certainly the ball will follow a parabolic path (projectile motion). So we need to know one about the angle of hitting.
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rakeen
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Re: Sewag dont hit too hard coz the meter is damaged!

Unread post by rakeen » Thu Apr 14, 2011 10:29 am

45 degree.
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raihatneloy
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Re: Sewag dont hit too hard coz the meter is damaged!

Unread post by raihatneloy » Fri Oct 28, 2011 3:30 pm

I am giving the solution below:
Image

Is it correct?

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Re: Sewag dont hit too hard coz the meter is damaged!

Unread post by raihatneloy » Fri Oct 28, 2011 3:31 pm

I am giving the solution below:
Image

Is it correct?

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Re: Sewag dont hit too hard coz the meter is damaged!

Unread post by raihatneloy » Sat Oct 29, 2011 11:39 am

Here, the ball will go to a distance of 102m if it is the maximum range of the motion!
so, here, \[R_{max}=102m\]
Again, we know,
\[R_{max}=\frac{u^2}{g}\]
or, \[102m=\frac{u^2}{g}\]
or, \[u=\sqrt{102\times 9.8}
=31.6165ms^{-1}\] ( Ans )

Is it correct? ? ? ? ? ? ?

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