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Q. 4.32  (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by   \theta (t)=tan^{-1}\left [ \frac{v_{0y-gt}}{v_{0x}} \right ]

where the symbols have their usual meaning.

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Using the equation of motion in both horizontal and vertical direction.

                                         v_y\ =\ v_{oy}\ =\ gt         and        v_x\ =\ v_{ox}

Now,                                

                                                 \tan \Theta \ =\ \frac{v_y}{v_x}

or                                                             =\ \frac{v_{oy}\ -\ gt}{v_{ox}}

Thus,                                              \Theta \ =\ \tan^{-1} \left ( \frac{v_{oy}\ -\ gt }{v_{ox}} \right )

Posted by

Devendra Khairwa

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