Sewag dont hit too hard coz the meter is damaged!
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 .
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Re: Sewag dont hit too hard coz the meter is damaged!
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.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
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Re: Sewag dont hit too hard coz the meter is damaged!
I am giving the solution below:
Is it correct?
Is it correct?
-
- Posts:3
- Joined:Fri Oct 28, 2011 3:10 pm
Re: Sewag dont hit too hard coz the meter is damaged!
I am giving the solution below:
Is it correct?
Is it correct?
-
- Posts:3
- Joined:Fri Oct 28, 2011 3:10 pm
Re: Sewag dont hit too hard coz the meter is damaged!
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? ? ? ? ? ? ?
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? ? ? ? ? ? ?