A projectile is projected with velocity v =3i + 4j .what is the ratio of max to min velocity ?
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Explanation:
ANSWER
Velocity of projectile v=3
i
^
+10
j
^
m/s
Speed of projectile u=∣v∣=
10
2
+3
2
=
109
m/s
Angle of projectile tanθ=
3
10
⟹ sinθ=
109
10
and cosθ=
109
3
Maximum height attained H=
2g
u
2
sin
2
θ
∴ H=
2(10)
109×
109
100
=5m
Horizontal range R=
g
u
2
sin2θ
=
g
2u
2
sinθcosθ
⟹ R=
10
2(109)×
109
10
×
109
3
=6m
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