Artillery Projectile

Posted on June 3rd, 2009 by admin

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Artillery Projectile

Physics - Projectile motion?

An artillery shell is fired at an angle of 46.2◦
above the horizontal ground with an initial
speed of 1950 m/s.
The acceleration of gravity is 9.8 m/s2 .
a)Find the total time of flight of the shell,
neglecting air resistance. Answer in units of
min.

b)Find its horizontal range, neglecting air resis-
tance. Answer in units of km.

The total flight time is twice the time it takes the shell to reach max altitutde since ballistic trajectories are symmetrical (the same going down as when they went up). The appropriate formula is the one for vertical velocity as a function of time after firing:

v = v0 - gt

v0 = initial vertical velocity = 1950sin(46.2) = 1294.147 m/s
g = acceleration of gravity = 9.8 m/s^2
v = 0 at max height since this is the point where the shell quits moving up and begins to fall down

0 = v0 - gt
t = v0/g = 1294.147/9.8 = 132.06 seconds

total flight time = 2(132.06) = 264.12 seconds = 4.402 minutes

The horizontal range is just the flight time times the initial horizontal velocity.

horizontal velocty = 1950cos(46.2) = 1458.66 m/s
range = (1458.66)(264.12) = 385260.8 m = 385.26 km

Seems like quite a long range for an artillery shell

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