A ball of mass 'm' is projected from the ground with a speed 'u' at an angle 'z' with the horizontal.The magnitude of the change in momentum of the ball over a time interval from beginning till it strikes the ground again is.

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Solution

$W e k n o w v e l o c i t y w i t h w h i c h b a l l s t r i k e t o t h e g r o u n d i s s a m e i n m a g n i t u d e a s t h e v e l o c i t y w i t h w h i c h b a l l i s p r o j e c t e d . i n i t i a l v e l o c i t y i n x d i r e c t i o n = u \cos z i n i t i a l v e l o c i t y i n y d i r e c t i o n = u \sin z f i n a l v e l o c i t y i n x d i r e c t i o n = u \cos z f i n a l v e l o c i t y i n y d i r e c t i o n : t i s t h e t i m e t a k e n t o r e a c h t h e h i g h e s t p o i n t . a t h i g h e s t p o i n t v e l o c i t y i s z e r o . 0 = u \sin z - g t t = \frac{u \sin z}{g} v e l o c i t y w i t h w h i c h i t s t r i k e t o t h e g r o u n d . v = g t = \frac{u \sin z}{g} g = u \sin z i n d o w n w a r d d i r e c t i o n c h a n g e i n v e l o c i t y i n x d i r e c t i o n = u \cos z - u \cos z = 0 c h a n g e i n v e l o c i t y i n y d i r e c t i o n = u \sin z - (- u \sin z) = 2 u \sin z c h a n g e i n m o m e n t u m i n = 2 m u \sin z$

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