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Q- A fielder whose height is 5 feet throw a ball with the initial velocity 25 m/s at 45 degree. The ball is again come down on the ground in 15 seconds. Find (i) the max distance covered (ii) the max height gain by the ball? 
Asked by sehgalgayrav810 | 18 Jun, 2020, 10:43: AM
answered-by-expert Expert Answer
M a x i m u m space d i s tan c e space c o v e r e d space i s space e q u a l space t o space t h e space r a n g e R a n g e equals V subscript H cross times T R a n g e equals 25 C o s 45 to the power of 0 cross times 15 R a n g e equals 25 cross times fraction numerator 1 over denominator square root of 2 end fraction cross times 15 R a n g e equals fraction numerator 375 over denominator square root of 2 end fraction m A t space M a x i m u m space h e i g h t space t h e space v e l o c i t y space w i l l space b e c o m e space z e r o s o space u sin g space 3 r d space e q u a t i o n space o f m o t i o n v squared equals u squared plus 2 a s 0 squared equals open parentheses 25 S i n 45 to the power of 0 close parentheses squared minus 2 g h h equals fraction numerator open parentheses 25 S i n 45 to the power of 0 close parentheses squared over denominator 2 g end fraction h equals fraction numerator 625 cross times begin display style 1 half end style over denominator 20 end fraction m M a x i m u m space h e i g h t space a c h i e v e d space b y space t h e space b a l l equals space 625 over 40 cross times 3.28 plus 5
1m=3.28ft
Answered by Utkarsh Lokhande | 18 Jun, 2020, 03:25: PM

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