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Question

# A body is thrown with velocity of 40 m/s in a direction making an angle of 300 with the horizontal. Calculate i) Horizontal range, ii) Maximum height and iii) Time taken to reach the maximum height.

A
(a)131.4m
(b)20.41m
(c)2.041s
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B
(a)121.4m
(b)24.41m
(c)2.091s
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C
(a)141.4m
(b)20.41m
(c)2.041s
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D
(a)161.4m
(b)10.41m
(c)2.051s
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Solution

## The correct option is C (a)141.4m(b)20.41m(c)2.041sR.E.F image u=40m/sθ=30∘ g=9.81Time of flight, T=2usinθgMaximum Height, H=u2sin2θ2gHorizontal Range, R=u2sin2θgTime taken to reach the max.height, t =T24=(40)2(sin30∘)22×9.8=20.408m=20.41mR=(40)2sin2×30∘9.8=141.4mt=I2=usin3θg=2.041sAns =141.4m,20.41m,2.041s (C)

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