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Application of Projectile to a Height

A body is pro...

Question

A body is projected with a velocity $u$ at an angle $θ$ with the horizontal. The velocity of the body will become perpendicular to the velocity of projection after a time $t$ given by.

\frac{2 u sin θ}{g}

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\frac{u sin θ}{g}

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\frac{2 u}{g sin θ}

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\frac{u}{g sin θ}

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Solution

The correct option is D $\frac{u}{g sin θ}$
$\begin{matrix} v_{y} = u sin θ - g t v_{x} = u cos θ \Rightarrow (u cos θ^j + u sin θ^j) \cdot (u cos θ^i + (u sin θ - g t)^j) = 0 \Rightarrow u^{2} {cos}^{2} θ + u^{2} {sin}^{2} θ - u sin θ g t = 0 \Leftarrow u^{2} = u sin θ g t \Rightarrow t = \frac{u}{g sin θ} \Rightarrow H e n c e, o p t i o n D i s c o r r e c t a n s w e r . \end{matrix}$

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A body is projected with a velocity u at an angle θ with the horizontal. The velocity of the body will become perpendicular to the velocity of projection after a time t given by.

The correct option is D ugsinθvy=usinθ−gtvx=ucosθ⇒(ucosθ^j+usinθ^j)⋅(ucosθ^i+(usinθ−gt)^j)=0⇒u2cos2θ+u2sin2θ−usinθgt=0⇐u2=usinθgt⇒t=ugsinθ⇒Hence,optionDiscorrectanswer.

A body is projected with a velocity $u$ at an angle $θ$ with the horizontal. The velocity of the body will become perpendicular to the velocity of projection after a time $t$ given by.