Solution.
x = 86 t ; x = 86t; x = 86 t ; y = 96 t − 4.9 t 2 ; y = 96t-4.9t^2; y = 96 t − 4.9 t 2 ;
Since the trajectory of motion is a parabola, the motion of the projectile is described by the formulas:
x = υ 0 x = \upsilon_0 x = υ 0 x t ; t; t ; y = υ 0 y = \upsilon_0 y = υ 0 y t + a y t 2 / 2 ; t + a_yt^2/2; t + a y t 2 /2 ;
a) a x = 0 ; a y = − 9.8 m / s 2 ; a_x = 0; a_y = -9.8m/s^2; a x = 0 ; a y = − 9.8 m / s 2 ;
b) υ x = 86 m / s ; \upsilon_x = 86m/s ; υ x = 86 m / s ; υ y = 96 − 9.8 t ; \upsilon_y = 96-9.8t; υ y = 96 − 9.8 t ;
c) υ 0 = υ x 2 + υ y 2 ; \upsilon_0 =\sqrt{\upsilon_x^2 + \upsilon_y^2 }; υ 0 = υ x 2 + υ y 2 ; υ 0 = 8 6 2 + 9 6 2 = 128.9 m / s ; \upsilon_0 = \sqrt{86^2 + 96^2 } = 128.9 m/s; υ 0 = 8 6 2 + 9 6 2 = 128.9 m / s ;
d) υ x = 86 m / s ; υ y = 96 − 9.8 ⋅ 5 = 47 m / s ; \upsilon_x = 86 m/s; \upsilon_y = 96 - 9.8\sdot 5 = 47m/s; υ x = 86 m / s ; υ y = 96 − 9.8 ⋅ 5 = 47 m / s ;
e) h m = − υ 0 h_m = -\upsilon_0 h m = − υ 0 y 2 /2⋅ a y ; \sdot a_y; ⋅ a y ;
h m = − 9 6 2 / 2 ⋅ ( − 9.8 ) = 470.2 m ; h_m = -96^2/2\sdot(-9.8) = 470.2m; h m = − 9 6 2 /2 ⋅ ( − 9.8 ) = 470.2 m ;
f) L = υ 0 L =υ_0
L = υ 0 x t m ; t_m; t m ; t m = 2 ⋅ ( − υ t_m =2\sdot(-\upsilon t m = 2 ⋅ ( − υ 0y / a y ) ; /a_y); / a y ) ; t m = 2 ⋅ ( − 96 / − 9.8 ) = 19.6 s ; t_m = 2\sdot(-96/-9.8) =19.6s; t m = 2 ⋅ ( − 96/ − 9.8 ) = 19.6 s ;
L = 86 ⋅ 19.6 = 1684.9 m ; L = 86\sdot19.6 = 1684.9m; L = 86 ⋅ 19.6 = 1684.9 m ;
Answer: a) a x = 0 ; a y = − 9.8 m / s 2 ; a_x = 0; a_y = -9.8 m/s^2; a x = 0 ; a y = − 9.8 m / s 2 ;
b) υ x = 86 m / s ; \upsilon_x = 86m/s ; υ x = 86 m / s ; υ y = 96 − 9.8 t ; \upsilon_y = 96-9.8t; υ y = 96 − 9.8 t ;
c) υ 0 = 128.9 m / s ; \upsilon_0 = 128.9m/s; υ 0 = 128.9 m / s ;
d) υ x = 86 m / s ; υ y = 47 m / s ; \upsilon_x = 86 m/s; \upsilon_y = 47m/s; υ x = 86 m / s ; υ y = 47 m / s ;
e)h m = 470.2 m ; h_m = 470.2m; h m = 470.2 m ;
f)L = 1684.9 m ; L = 1684.9m; L = 1684.9 m ;
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