Question

13. A body is projected at an angle of
projection 35º. To get the same range
with the same velocity of projection, the
body should be projected at an angle
(a) 70° (b) 55° (c) 65
(d) 53​

Answer 1

Answer:

70

Explanation:

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Answer 2

Explanation:

we know ,

horizontal range = u²sin2Ф/g

maximum height = u²sin²Ф/2g

A/C to question,

horizontal range = 3× maximum height

u²sin2Ф/g = 3 × u² sin²Ф/2g

{ sin2Ф = 2sinФ.cosФ }

2sinФ.cosФ = 3/2. sin²Ф

tanФ = 4/3

Ф = tan⁻¹(4/3)