Q126 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Physics

The surface of a hill inclined at an angle 30° to the horizontal. A stone is thrown from the summit of the hill (point A) at an initial speed 10 m/s at angle 60° to the vertical. If the stone strikes the hill at point B as shown in the figure, the distance between A and B is (Take g = 10 m/s2) Figure

  • A
    $10 \mathrm{~m}$
  • B
    $15 \mathrm{~m}$
  • Correct answer
    C
    $20 \mathrm{~m}$
  • D
    $30 \mathrm{~m}$

Solution

Component of velocity along incline $=\left(10 \cos 60^{\circ}\right)=5 \mathrm{~m} / \mathrm{s}$ and, $\mathrm{T}=\frac{2\left(10 \sin 60^{\circ}\right)}{\mathrm{g} \cos 30^{\circ}}=2 \mathrm{sec} \quad\left[\because \mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g} \cos \alpha}\right]$ $$ \begin{aligned} & \text { So, } \mathrm{AB}=5 \times 2+\frac{1}{2} \times \mathrm{g} \sin 30^{\circ} \times 2^2 \\ & =10+10=20 \mathrm{~m} \end{aligned} $$

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

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