Question

An object shot into the air follows the path given by F(t) = (at, bt – 4.9t2) m with t in seconds and a and b are unknown physical constants. y (m) 500 m/s 30° (m) The launch velocity is 500 m/s at an angle of 30°. What is the total distance traveled through the air? Be accurate to within 10 meters. 22092.5 m x

Answer 1

Solna Îlt) lat, bt- 4.94 ) g/m2 500 singo / room's mo 300 toim) 250 ml 500001 30² m/s & 25053 m/s so have we a=25063 mls 250 ml and ba <25053 +, 250 7-4.9t's nit) for time se 2 2500 49 sel anc Elt TUS flight y+) 2.0 250t- utgq² = 0 = +(25o- ungt) = 0 ta 250 t20 4.9 Now, the total distance traveled through air will be equal to length Arce length Sveppe de : *3 254 212 250537 2500/4g at Vesos) (250–9.8474842 : 2104-4.942 & offus +8800-4900€ let 250-9.87 25053 tann dta : 25053 2 is dy 6 Tag 1 . 250 -9.87 12 6 ) g. 8x 250 13 stu?nda

when tzo tanaa % Ja when t= 2500 Jamya 26053 Janna (250-9.88 2500 ething (200-500) 3) La $32-176 2053 substituting 6 L: -) 12:053), S i + fas? x 25053 see? ada X 9.8 176 -) La ņu esoy?x3see? Podx - rl 9.8 r/6 seex fandtln/ sex+fanal, siera d) orice ala (250) 2 x3 ·9.8*2 % [. (20) ²x3 9.6x2 seqgle landeng 12 [ se se q fonti - self-9 3 tank 9]+ ln { sue que donen şi la (25092x3 (250) ²73 3 몰 => L = 123264.78 prima stagione 279.8 ܕ ܐ ܙ м.