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Motion in a Plane

Section: Physics

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Question : 27 of 28

Marks: +1, -0

One second after projection, a projectile is travelling in a direction inclined at

45^{\circ}

to horizontal. After two more seconds it is travelling horizontally. Then the magnitude of velocity of the projectile is

(g=10\ \mathrm{m}\,\mathrm{s}^{-2})

[AP EAPCET 18 May 2024 Shift 1]

$10 \sqrt{13}\ \mathrm{m}\,\mathrm{s}^{-1}$
$11\ \mathrm{m}\,\mathrm{s}^{-1}$
$10 \sqrt{2}\ \mathrm{m}\,\mathrm{s}^{-1}$
$20\ \mathrm{m}\,\mathrm{s}^{-1}$

Solution:

Let in is body reaches upto point

A

and after two more second, it reaches upto point

B

Total time of ascent for a body is,

t_{\;\text{ascent }\;}=2+1=3\ \text{s}

t_{\;\text{ascent }\;}= \frac{\;\text{ Time of flight }\;}{2}= \frac{2 u \sin \theta}{2 \times g}

\frac{u \sin \theta}{g}=3 \Rightarrow u \sin 45^{\circ}=30 \dots

Horizontal component of velocity remains always constant,

u \cos 45^{\circ}=v \cos 45^{\circ}

For vertical upward motion between point

O

and

A

,

v_y=u_y-g t

v \sin 45^{\circ}=u \sin 45^{\circ}-10 \times 1

\frac{v}{\sqrt{2}}=30-10 \quad \text{ from Eq. (i) }

v=20 \sqrt{2}\ \text{m/s}

Putting this value in Eq. (ii),

u \cos 45^{\circ}=20 \sqrt{2} \times \frac{1}{\sqrt{2}}=20

From Eq. (i) and Eq. (iii),

u^2 \sin^2 45^{\circ}+u^2 \cos^2 45^{\circ}=(30)^2+(20)^2

u^2=1300

u=10 \sqrt{13}\ \text{m/s}

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