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Q. 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 $\hat{ i }, \hat{ j }$ 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 $\left(g=10\, ms ^{-2}\right)$

AP EAMCETAP EAMCET 2017

Solution:

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Time in which body is dropped to ground is calculated using following data;
$u=5\, m / s$
(k is given vertically upward direction)
$a=-10\, m / s ^{2}$
$h=-30\, m$
Now using the formula
$h=u t+\frac{1}{2} a t^{2}$
$-30=5 t-\frac{1}{2} \times 10 \times t^{2}$
$\Rightarrow t^{2}-t-6=0$
$\Rightarrow t=-2 s$ (Not possible)
or $t=3\, s$
In this time, projectile moving in East and North with speeds $3 m / s$ and $4 m / s$. Distances covered in these directions are;
In east $(x-$ coordinate).
- speed $\times \text { time }=3 \times 3=9\, m$
and in North ( $y$ - coordinate)
$=4 \times 3=12\, m$
So, Projective lando at $(x, y)=(9 m, 12 m)$ mark.
So, horizontal range of body on ground is;
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Range $=\sqrt{9^{2}+12^{2}}$
$=\sqrt{225}$
$=15\, m$