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A particle is thrown with 10m/s at an angle of 60^{\circ} with horizontal the time at which its velocity is perpendicular to the initial velocity is ( g = 10 m/s )

Option: 1

\sqrt{3}\; sec


Option: 2

2\sqrt{3}\; sec


Option: 3

\frac{2}{\sqrt{3}}\; sec


Option: 4

\frac{4}{\sqrt{3}}\; sec


Answers (1)

best_answer

As we learned

Scalar , Dot or Inner Product -

Scalar product of two vector \vec{A} & \vec{B} written as \vec{A} \cdot \vec{B} is a scalar quantity given by the product of magnitude of \vec{A} & \vec{B} and the cosine of smaller angle between them.

\vec{A}\cdot \vec{B}= A\, B\cdot \cos \Theta

- wherein

showing representation of scalar products of vectors.

 

 

\\*u=10\cos 60^{\circ}\hat{i}+10\sin 60^{\circ}\hat{j}\\*\\*=5(\hat{i}+\sqrt{3}\hat{j})

 \\*\\*V_{y}=u_y-gt=5\sqrt{3}-10t(here \ use \ g=10)\\*\\*\vec{V}=5\hat{i}+(5\sqrt{3}-10t)\hat{j}\\*\\*since\; \vec{V}\; is\; perpendicular\; to\; \vec{u}hence\; \vec{u}\cdot \vec{V}=0\\*\\*\Rightarrow 5(\hat{i}+\sqrt{3}\hat{j})\cdot 5(\hat{i}+(\sqrt{3}-2t)\hat{j})=0\\*\\*\Rightarrow 1+\sqrt{3}(\sqrt{3}-2t)=0

\\*or\; 1+3-2\sqrt{3}t=0\\*\\*\Rightarrow t=\frac{2}{\sqrt{3}}sec

 

Posted by

Deependra Verma

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