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Q - A ball is launched with initial velocity 25 m/s at 45 degree by a ball launcher. The initial height is 0. The ball come back on ground in 20 seconds. Find (i) the max distance covered (ii) the max height gain by the ball?
Asked by sehgalgaurav810 | 18 Jun, 2020, 11:16: 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 20 R a n g e equals 25 cross times fraction numerator 1 over denominator square root of 2 end fraction cross times 20 R a n g e equals 250 square root of 2 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
Answered by Utkarsh Lokhande | 18 Jun, 2020, 03:18: PM

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