A body is projected from ground with velocity u→=(40i^+30j^) m/s. Then the ratio of its range to the maximum height attained is

ucos⁡α=40 and usin⁡α=30Now, RH=(u2gsin⁡2α)(u2sin⁡2α2g)=4cot⁡α=4×4030=163

A body is projected from ground with velocity u→=(40i^+30j^) m/s. Then the ratio of its range to the maximum height attained is

ucos⁡α=40 and usin⁡α=30Now, RH=(u2gsin⁡2α)(u2sin⁡2α2g)=4cot⁡α=4×4030=163

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A body is projected from ground with velocity $\vec{u} = (40 \hat{i} + 30 \hat{j}) m / s$ . Then the ratio of its range to the maximum height attained is

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9 : 4

15 : 8

16 : 3

12 : 5

answer is C.

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$\begin{array}{l} ucos α = 40 and usin α = 30 \\ Now, \frac{R}{H} = \frac{(\frac{u^{2}}{g} \sin 2 α)}{(\frac{u^{2} \sin^{2} α}{2 g})} = 4 \cot α = 4 \times \frac{40}{30} = \frac{16}{3} \end{array}$

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A body is projected from ground with velocity u→=(40i^+30j^) m/s. Then the ratio of its range to the maximum height attained is

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A body is projected from ground with velocity $\vec{u} = (40 \hat{i} + 30 \hat{j}) m / s$ . Then the ratio of its range to the maximum height attained is