An artillery shell is launched on a flat, horizontal field at an angle of α = 40.5° with respect to the horizontal and with an initial speed of v0 = 338 m/s.

Question

a.What is the horizontal distance covered by the shell after 8.09 s of flight?
b.What is the height of the shell at this moment?

Answer 1

horizontal problem:
u = 338 cos 40.5 = 257 m/s

x = u t = 257 * 8.09

h = Vi t - 4.9 t^2
where Vi = 338 sin 40.5 = 220 m/s
h = 220(8.09) - 4.9 (8.09)^2