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JEE-ADV PHYSICS PRACTICE EXERCISE VOLI
40. A body is projected horizontally from
47. A
certain height. After 13 sec, its direction
of motion make 60 with the horizontal
Then its initial velocity of projection is
1)9.8m/s 234.9m/s 3)19.6m/s 4) 14.7m/s
6
Tm
Answers
Answered by
0
x=40cosθ×2=80cosθ
Applying equation of trajectory ,
20=80cosθ×tanθ−
2×cos
2
θ×40
2
g(80cosθ)
2
20=80sinθ−
2
40
sinθ=
2
1
,θ=30
∘
Range =
g
u
2
sin2θ
=
20
1600×
3
=80
3
=138.5m
Answered by
0
Answer:
x=40cosθ×2=80cosθ
Applying equation of trajectory ,
20=80cosθ×tanθ−
2×cos
2
θ×40
2
g(80cosθ)
2
20=80sinθ−
2
40
sinθ=
2
1
,θ=30
∘
Range =
g
u
2
sin2θ
=
20
1600×
3
=80
3
=138.5m
Explanation:
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