t = 6.77 s
X = Vx*t , where
Vx = Vo*cos43.5 = 190.8 m/s
Y = Vyo*t - (g/2) t^2, where
Vyo = Vo*sin43.5 = 181.04 m/s and
g = 9.81 m/s^2
X and Y are your two answers. Proceed with the calculation.
An artillery shell is launched on a flat, horizontal field at an angle of a = 43.5° with respect to the horizontal and with an initial speed of v0 = 263 m/s. What is the horizontal distance covered by the shell after 6.77 s of flight?
What is the height of the shell at this moment?
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