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A projectile is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal direction with a velocity of $147 \mathrm{~ms}^{-1}$. Then the time after which its inclination with the horizontal is $45^{\circ}$, is
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$5.49 s$
Key Idea : At the two points of the trajectory during projection, the horizontal component of the velocity is the same.
Horizontal component of velocity at angle $60^{\circ}$
$=$ Horizontal component of velocity at $45^{\circ}$
$i e, \quad u \cos 60^{\circ}=v \sin 45^{\circ}$
or $\quad 147 \times \frac{1}{2}=v \times \frac{1}{\sqrt{2}}$
or $\quad v=\frac{147}{\sqrt{2}} \mathrm{~m} / \mathrm{s}$
Vertical component of $u=u \sin 60^{\circ}$
$=\frac{147 \sqrt{3}}{2} \mathrm{~m}$
Vertical component of $v=v \sin 45^{\circ}$
$=\frac{147}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{147}{2} \mathrm{~m}$
but $\quad v_y=u_y+a t$
$\therefore \quad \frac{147}{2}=\frac{147 \sqrt{3}}{2}-9.8 t$
or $\quad 9.8 t=\frac{147}{2}(\sqrt{3}-1)$
$\therefore \quad t=5.49 \mathrm{~s}$
Horizontal component of velocity at angle $60^{\circ}$
$=$ Horizontal component of velocity at $45^{\circ}$
$i e, \quad u \cos 60^{\circ}=v \sin 45^{\circ}$
or $\quad 147 \times \frac{1}{2}=v \times \frac{1}{\sqrt{2}}$
or $\quad v=\frac{147}{\sqrt{2}} \mathrm{~m} / \mathrm{s}$
Vertical component of $u=u \sin 60^{\circ}$
$=\frac{147 \sqrt{3}}{2} \mathrm{~m}$
Vertical component of $v=v \sin 45^{\circ}$
$=\frac{147}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{147}{2} \mathrm{~m}$
but $\quad v_y=u_y+a t$
$\therefore \quad \frac{147}{2}=\frac{147 \sqrt{3}}{2}-9.8 t$
or $\quad 9.8 t=\frac{147}{2}(\sqrt{3}-1)$
$\therefore \quad t=5.49 \mathrm{~s}$
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