Answer:
a) Shell's horizontal range = 243798.70 m
b) The amount of time the shell is in motion = 197.06 s
Explanation:
Consider the vertical motion of projectile,
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 1.57 x 10³ sin 38 = 966.59 m/s
Acceleration, a = -9.81 m/s²
Displacement is zero when the shell reaches back to ground, s = 0 m
Time, t =?
Substituting,
s = ut + 0.5 at²
0 = 966.59 x t + 0.5 x (-9.81) x t²
t = 197.06 seconds
b) The amount of time the shell is in motion = 197.06 s
a) Consider the horizontal motion of shell
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 1.57 x 10³ cos 38 = 1237.18 m/s
Acceleration, a = 0 m/s²
Displacement, s = ?
Time, t = 197.06 s
Substituting,
s = ut + 0.5 at²
s = 1237.18 x 197.06 + 0.5 x 0 x 197.06²
s =243798.70 m
Shell's horizontal range = 243798.70 m
