horizontal problem:
u = 1600 cos 43 the whole time
range = u t
where t is the time in the air
vertical problem to get t:
initial vertical speed = Vi = 1600 sin 43
v = Vi - g * time
at top, v = 0 and time = t/2
so
0 = 1600sin 43 - 9.81 (t/2)
so
t = (3200 sin 43 /9.81)
now go back and use that t in range = u t
A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an initial angle of 43.0◦ to the horizontal.
The acceleration of gravity is 9.81 m/s2 .
a) Neglecting air resistance, find the shell’s horizontal range.
Answer in units of m.
b) How long is the shell in motion? Answer in units of s.
1 answer