Projection Under uniform Acceleration

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Question

A body is projected at an angle 30° to the horizontal with a speed of 30 ms^–1. The angle made by the velocity vector with the horizontal after 1.5 s is (g = 10 ms^–2)

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Solution

$\large Tan\alpha = \frac{{{V_y}}}{{{V_x}}} = \frac{{u\sin \theta - gt}}{{u\cos \theta }} = \frac{{30\sin {{30}^0} - 10 \times 1.5}}{{30 \times \cos {{30}^0}}}$
$\large Tan\alpha = 0 \Rightarrow \alpha = {0^0}$

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Projection Under uniform Acceleration

Remember concepts with our Masterclasses.

A body is projected at an angle 30° to the horizontal with a speed of 30 ms–1. The angle made by the velocity vector with the horizontal after 1.5 s is (g = 10 ms–2)

Ready to Test Your Skills?

free study material

A body is projected at an angle 30° to the horizontal with a speed of 30 ms^–1. The angle made by the velocity vector with the horizontal after 1.5 s is (g = 10 ms^–2)