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Exercise 0 – Ideal ballistics
• Motion in vacuum is analytical:
v = v = v cosφ
x
0x
0 0
v = v − gt = v sinφ
−gt
y
0y
0 0
• The trajectory is a parabola:
φ = φ and v = v
2 0 2 0
• Time to reach highest point is when v = 0 , i.e.:
v0 y v0
t1 = = sinφ0
g g
y
• By symmetry, time to impact is twice as long:
v0 y v0
t2 = 2t1 = 2 = 2 sinφ0
g g
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Exercise 0 – Ideal ballistics (2)
• The rangeis therefore:
2 2
v0 y v0 v0
X = vxt2 = 2v0x
= 2 sinφ0cosφ0 = sin(2 φ0)
g g g
v = v = v cosφ
x
0x
0 0
v = v − gt = v sinφ
−gt
y
0y
0 0
v0 y v0
t1 = = sinφ0
g g
v0 y v0
t2 = 2t1 = 2 = 2 sinφ0
g g
• Max range is when shooting at 45º:
2
0
X
max
( φ0) = v for sin(2 φ0) = 1 ⇔ φ0
= π 4
g
• The position at the generic time is:
xt () = v t=
vtcosφ
t
t
yt () = v0yt− g = vt
0
sinφ0
−g
2 2
0x
0 0
2 2
• Max elevation depends only on v 0y :
2 2 2 2
v0y g v0y v0y
v0
2
Y = ymax = y( t1) = − = = sin φ
2
0
g 2 g 2g 2g
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