Q150 / 200 · JIPMER 2008 · Physics

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

  • A
    $15 s$
  • B
    $10.98 s$
  • Correct answer
    C
    $5.49 s$
  • D
    $2.745 s$

Solution

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 $\quad 9.8 t=\frac{147}{2}(\sqrt{3}-1)$ $\therefore \quad t=5.49 \mathrm{~s}$

Hence, the correct answer is option: 'C'.

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