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A body is projected from the top of a tower with a velocity $\bar{u} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} {ms}^{- 1}$ , where $i, j and \hat{k}$ are unit vectors along east, north and vertically upwards respectively. If the height of the tower is 30 m. horizontal range of the body on the ground is (g=10 ms^-2)

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15 m

25 m

9 m

12 m

answer is A.

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Detailed Solution

$\begin{array}{l} h = \frac{1}{2} {gt}^{2} - μt \Rightarrow t = 3 s \Rightarrow x = μ_{i} t = 3 \times 3 = 9 m \\ (μ = 5 {ms}^{- 1}) y = μ_{y} t = 4 \times 3 = 12 m \\ Range, R = \sqrt{x^{2} + y^{2}} = 15 m \end{array}$

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