Artillery Crew Gun
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Artillery Crew Gun

Finding the angle of elevation?
An artillery crew wants to shell a position on level ground 28 km away. If the gun has a muzzle velocity of 810 m/s , to what angle of elevation should the gun be raised in degrees?
You have :
-----------------------------------------------------
XF = 28,000 m.
V0 = 810.0 m. per sec.
VXF = VX0 = ( V0 ) ( cos B )
XF = ( V0 ) ( cos B ) ( tF )
VY0 = ( V0 ) ( sin B )
tF = [ 2 ] [ VY0 / g ] = [ 2 ] [ V0 ] [ sin B ] [ 1 / g ]
XF = [ V0 ] [ cos B ] [ 2 ] [ V0 ] [ sin B ] [ 1 / g ]
[ cos B ] [ sin B ] = [ 1/2 ] [ sin 2B ]
XF = [ VO ]^2 [ 2 / g ] [ 1/2 ] [ sin 2B ] = [ V0 ]^2 [ 1/g ] [ sin 2B ]
sin 2B = [XF ] [ [ g ] / [ V0 ]^2
sin 2B = [ 28,000 ] [ 9.8 ] / [ 810 ]^2
sin 2B = 0.4182
2B = 24.74 deg.
B = 12.37 deg.<------------------------------
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St.Agustine-1830's Seminole War Artillery crew firing cannon
Need help on a physics projectile motion problem, thanks?
An artillery crew wants to shell a position on level ground 35km away. If the gun has a muzzle velocity of 770m/s, to what angle of elevation should the gun be raised?
At the last part herein, there is a prederived formula to use for Range. But here is the concept.
..............._----_ ←at this point Vy = 0
.........._-‾..........‾-_..………..Vx = still constant
...... _-‾................‾-…………Y or H is max.
Vy.../......................................t = ½ T (half-flight)
'.↑../.........................
..|./.غ )...........'.........
..|/---→Vx________________ ground level Y = 0
▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓…
..|←-------35km.----------→|
X = 35km(1000m/1km) = 35000m
Vy = Vsin Ø = 770(sin Ø)
Vx = VcosØ = 770(cosØ)
X = Vx t
35000 = 770(cosØ)t
(cosØ) = (35000) / (770t)
cosØ = 45.45 / t .....................◄eq1
For the shell to return to the ground, Y = 0
S = Vit + ½ at ²
.↓ .....↓ .........↓
Y = Vyt + ½ (-g)t ²
0 = ( Vsin Ø )t - (4.9)t ²
(4.9)t ² = Vsin Ø )t ...........divide by t
4.9t = Vsin Ø
t = 770sin Ø / 4.9
t = (157.14)sin Ø ............◄eq 2
substitute t from ◄eq 2 to ◄eq1
cosØ = 45.45 / (157.14)sin Ø
(sin Ø)cos Ø = 0.289
..................in trigonometry
...................sin 2Ø = 2 sin Øcos Ø
....................(sin Ø)cos Ø = ½ sin 2 Ø
½ sin 2 Ø = 0.289
sin 2 Ø = 0.578
2 Ø = sin‾ ¹ (0.578)
2 Ø = 35.34º
Ø = 17.67º
To CHECK: Use the prederived formula for Range
Range = Vo²(sin2Ø)/g
35000 = (770)² sin2 Ø / (9.8)
sin2 Ø = 35000(9.8) / (770)²
.............= 0.578
2Ø = 35.34º
Ø =17.67º CHECK
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