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From the top of the building, two balls are thrown horizontally with velocities\mathrm{ u_1} and \mathrm{ u_2} in opposite directions. If their velocities are perpendicular to each other just before they strike the ground, The height of the building is-

Option: 1

\mathrm{\frac{u_1^2}{2 g}}


Option: 2

\mathrm{\frac{u_1 u_2}{2 g}}


Option: 3

\mathrm{\frac{u_2^2}{2 g}}


Option: 4

\mathrm{\frac{u_1 u_2}{g}}


Answers (1)

best_answer

Use Newton's Equation of motion -

\mathrm{ \vec{v}_1=u_1 \hat{i}-g t \hat{j} \text { \& } \vec{v}_2=-u_2 \hat{i}-g t \hat{j} }

Given, velocities are perpendicular to each other

             \mathrm{ \vec{v}_1 \cdot \vec{u}_2=0 }

or,        \mathrm{\left(u_1 \hat{i}-g t \hat{\jmath}\right) \cdot-\left(u_2 \hat{i}+g \hat{\jmath}\right)=0}

                                    \mathrm{ t=\frac{\sqrt{u_1 u_2}}{g} \quad\left[\begin{array}{c} \because \hat{\imath} \cdot \hat{\imath}=1 \\ \hat{\jmath} \cdot \hat{\jmath}=1 \end{array}\right] }

Now, 

\mathrm{\begin{aligned} h & =\frac{1}{2} g t^2 \\ & =\frac{1}{2} g \times\left(\frac{\sqrt{u_1 u_2}}{g}\right)^2 \\ h & =\frac{1}{2} g \times \frac{u_1 u_2}{g^2} \end{aligned}}

\mathrm{h=\frac{u_1 u_2}{2 g} \text { Ans. }}

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Rishi

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