fisica 04 05 soluzioni esercizi moto parabolico -2- 1 1

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x(t) = v0 · t
y(t) = 1.2 − 4.905 · t2
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q
1.2 ∼
tv
y(tv ) = 0 = 1.2 − 4.905 · t2v
⇒
tv =
4.905
= 0.495
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de
1.5 ∼
x(tv ) = v0 · tv = 1.5
⇒
v0 =
tv
= 3.03
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de
vx (tv ) = v0 ∼
= 3.03
O%
vy (tv ) =q9.81 · tv ∼
= 4.85
O%
2
2 ∼
v(tv ) =
vx + vy = 5.72
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) ∼ 58◦
α = tan−1 ( 4.85
3.03 =
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= 15 ms
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x(t) = 15 · cos(30◦ ) · t
y(t) = 8 + 15 · sin(30◦ ) · t − 4.905 · t2
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v
y(tv ) = 0 = 8 + 15 · sin(30◦ ) · tv − 4.905 · t2v ⇒ tv ∼
= 2.25
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◦
∼
x(tv ) = 15 · cos(30 ) · tv = 29.3
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ZN;2 t 8!"7;<#&&GgA <">%&H#`;Y t =
s
s
◦
2 ∼
y(ts ) = 8 + 15 · sin(30 ) · ts − 4.905 · ts = 10.9
15∗sin(30◦ )
9.81
A
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◦
∼
x(ts ) = 15 · cos(30 ) · ts = 9.93
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◦ ∼
vx (ts ) = v0x = 15 · cos(30 ) = 13.0
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◦ ∼
vx (tv ) = v0x = 15 · cos(30 ) = 13.0
d%
vy (tsq
) = v0y − 9.81 · tv = 15 · sin(30◦ ) − 9.81 · tv ∼
= −14.6
d%
v = vx2 + vy2 ∼
= 19.5
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α
α=
tan−1 ( vvxy )
∼
= 48.3◦
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v0
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α
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x(t) = v0 · cos(α) · t
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y(t) = v0 · sin(α) · t − 12 g · t2
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v0 ·sinα
ts tv = 2 ·
tv
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d = x(tv ) = v0 · cosα · 2 ·
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v0
f7
α
v0 ·sinα
g
= 2 · v02 ·
g
sinα·cosα
g
Fa] %&#D??H#QQGH7%:!;&&
h = y(ts ) = v0 · sinα ·
v0 ·sinα
g
− 21 g · ( v0 ·sinα
)2 =
g
1
2
· v02 ·
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d=3·h
⇒
sinα
cosα
⇒
=
4
3
2 · v02 ·
⇒
sinα·cosα
g
tan(α) =
=
4
3
3
2
· v02 ·
sin2 α
g
⇒
α = tan−1 ( 34 ) ∼
= 53.1◦
⇒
2 · cosα =
3
2
· sinα
sin2 α
g

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