A body is projected from the top of a tower with a velocity $\overrightarrow{\mathbf{u}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} \mathrm{ms}^{-1}$, where $\hat{\mathbf{i}}, \hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$ are unit vectors along east, north and vertically upwards…

Question

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

Aicharya · Accepted Answer

( $\hat{\mathbf{k}}$ is given vertically upward direction) $ \begin{aligned} & a=-10 \mathrm{~m} / \mathrm{s}^2 \\ & h=-30 \mathrm{~m} \end{aligned} $ As, $S=u t+\frac{1}{2} a t^2 \Rightarrow-30=5 t-\frac{1}{2} \times 10 \times t^2$ $ \Rightarrow t^2-t-6=0 $ $ \Rightarrow t=-2 s \text { (Not possible) } $ or, $t=32 \mathrm{~s}$. This is time in which body reaches the ground In this time, projectile moving in East and North with speeds $3 \mathrm{~m} / \mathrm{s}$ and $4 \mathrm{~m} / \mathrm{s}$. Distances covered in these directions are; In east $(x$ - coordinate $)=3 \times 3=9 \mathrm{~m}$ and in North $(y$ - coordinate $)=4 \times 3=12 \mathrm{~m}$ So, Projectile land at $(x, y) \equiv(9 \mathrm{~m}, 12 \mathrm{~m})$ mark. So, horizontal range of body on ground is:

Range $=\sqrt{9^2+12^2}=15 \mathrm{~m}$

A body is projected from the top of a tower with a velocity $\overrightarrow{\mathbf{u}}=3…

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