A stone projected at an angle of 60- from the ground level strikes at an angle of 30- on the roof of a building of height -h- Then the speed of projection of the stone is -2gh-3gh-gh-6gh-

The correct option is B √(3gh)tan 30^∘ = v_2y/v_x and tan 60^∘ = v_1y/v_x ∴ v_1y = 3×v_2y Now, v^2_1y v^2_2y = 2gh ⇒ v_2y = √(gh/4) or v_x = √(3gh/4) or v ...

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Byju's Answer

Standard XII

Physics

Range

A stone proje...

Question

A stone projected at an angle of $60^{\circ}$ from the ground level strikes at an angle of $30^{\circ}$ on the roof of a building of height ‘h’. Then the speed of projection of the stone is

\sqrt{2 g h}

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\sqrt{6 g h}

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\sqrt{3 g h}

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\sqrt{g h}

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Solution

The correct option is C $\sqrt{3 g h}$
$t a n 30^{\circ} = \frac{v_{2 y}}{v_{x}} a n d t a n 60^{\circ} = \frac{v_{1 y}}{v_{x}} ∴ v_{1 y} = 3 \times v_{2 y} N o w, v_{1 y}^{2} - v_{2 y}^{2} = 2 g h \Rightarrow v_{2 y} = \sqrt{\frac{g h}{4}} o r v_{x} = \sqrt{\frac{3 g h}{4}} o r v = \frac{v_{x}}{c o s 60^{\circ}} = \sqrt{3 g h}$

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A stone projected at an angle of 60∘ from the ground level strikes at an angle of 30∘ on the roof of a building of height ‘h’. Then the speed of projection of the stone is

The correct option is C √3ghtan 30∘=v2yvxand tan 60∘=v1yvx∴v1y=3×v2yNow,v21y−v22y=2gh⇒v2y=√gh4or vx=√3gh4or v=vxcos 60∘=√3gh

A stone projected at an angle of $60^{\circ}$ from the ground level strikes at an angle of $30^{\circ}$ on the roof of a building of height ‘h’. Then the speed of projection of the stone is