u = 1540 cos 52 = 948 the whole time
Vi = 1540 sin 52 = 1214 at start
v = Vi - 9.8 t
0 = 1214 - 9.8 t
t = 124 seconds to top
so
2 t = 248 seconds in the air (part B)
range = u * 2t = 234,871 meters or 235 km
A shell is fired from the ground with an initial speed of 1.54 ✕ 103 m/s at an initial angle of 52° to the horizontal.
(a) Neglecting air resistance, find the shell's horizontal range.
(b) Find the amount of time the shell is in motion.
2 answers
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