A ball is thrown in vertical direction with a speed of 25 m/s. At the same instant another ball is dropped from the height of 100 m. Find when and where the balls will meet.

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Solution

$H e i g h t c o v e r e d b y t h e b a l l f r o m g r o u n d h = u t - \frac{1}{2} g t^{2} h = 25 t - \frac{1}{2} \times 10 t^{2} h = 25 t - 5 t^{2} . . . . . . . . . . . . . . . . . . . (1) d i s \tan c e c o v e r e d b y t h e b a l l t h r o w n d o w n i n t i m e t w i l l b e : \sin c e i t f a l l s f r o m r e s t, u = 0 100 - h = \frac{1}{2} g t^{2} 100 - (25 t - 5 t^{2}) = 5 t^{2} . . . . . . . . . . (f r o m 1) 25 t = 100 t = 4 s T h u s t h e b a l l m e e t e a c h o t h e r a t 4 s H e i g h t a t w h i c h t h e b a l l m e e t i s h h = 25 t - 5 t^{2} h = 25 \times 4 - 5 {(4)}^{2} h = 100 - 80 = 20 m$

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