An artillery shell is launched on a flat, horizontal field at an angle of α = 42.6° with respect to the horizontal and with an initial speed of v0 = 327 m/s.
a) What is the horizontal distance covered by the shell after 3.73 s of flight?
b) What is the height of the shell at this moment?
Gravitational acceleration = g = -9.81 m/s2
Initial velocity of the artillery shell = V0 = 327 m/s
Angle of launch = =
42.6o
Initial horizontal velocity of the artillery shell = Vx0
Vx0 = V0Cos
Vx0 = (327)Cos(42.6)
Vx0 = 240.704 m/s
Initial vertical velocity of the artillery shell = Vy0
Vy0 = V0Sin
Vy0 = (327)Sin(42.6)
Vy0 = 221.338 m/s
Time period = T = 3.73 s
There is no horizontal force acting on the shell hence the velocity of the shell in the horizontal direction doesn't change.
Horizontal distance covered by the shell in 3.73 sec = R
R = Vx0T
R = (240.704)(3.73)
R = 897.83 m
Height of the shell at 3.73 sec = H
H = Vy0T + gT2/2
H = (221.338)(3.73) + (-9.81)(3.73)2/2
H = 757.35 m
a) Horizontal distance covered by the shell after 3.73 sec of flight = 897.83 m
b) Height of the shell at this moment = 757.35 m
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