A body is projected at t = 0 with a velocity 10 ms⁻¹ at an angle of 60º with the horizontal. The radius of curvature of its trajectory at t = 1s is R. Neglecting air resistance and taking acceleration due to gravity g = 10 ms⁻², the radius of R is:
(A) 10.3 m (B) 2.8 m
(C) 2.5 m (D) 5.1 m
Answers
Answered by
2
vx=10cos60o=5m/s
vy=10cos30o=53m/s
Velocity after t=1 sec
vx=5m/s
vy=∣(53−10)∣m/s=10−53
an=Rv2⇒anvx2+vy2=10cosθ25+100+75−1003
tanθ=10−5root3=2−root ⇒θ=15o
R=100(2−root3)/10cos15=2.8m
Answered by
2
v
x
=10cos60
o
=5m/s
v
y
=10cos30
o
=5
3
m/s
Velocity after t=1 sec
v
x
=5m/s
v
y
=∣(5
3
−10)∣m/s=10−5
3
a
n
=
R
v
2
⇒
a
n
v
x
2
+v
y
2
=
10cosθ
25+100+75−100
3
tanθ=
5
10−5
3
=2−
3
⇒θ=15
o
R=
10cos15
100(2−
3
)
=2.8m
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