Projection Under uniform Acceleration

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Question

Two seconds after projection, a projectile is moving at 30° above the horizontal. After one more second it is moving horizontally. If g = 10ms^-2, the velocity of projection is

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Solution

$\large Tan\alpha = \frac{{u\sin \theta - gt}}{{u\cos \theta }}$
$\large Tan{30^0} = \frac{{u\sin \theta - 2g}}{{u\cos \theta }}$
$\large \tan 0 = \frac{{u\sin \theta - 3g}}{{u\cos \theta }}$
From the above equation
$\large \left. {\begin{array}{*{20}{c}} {u\sin \theta = 3g}\\ {u\cos \theta = \sqrt 3 g} \end{array}} \right\} \Rightarrow U = 20\sqrt 3 \,m/s$

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Projection Under uniform Acceleration

Remember concepts with our Masterclasses.

Two seconds after projection, a projectile is moving at 30° above the horizontal. After one more second it is moving horizontally. If g = 10ms-2, the velocity of projection is

From the above equation

Ready to Test Your Skills?

free study material

Two seconds after projection, a projectile is moving at 30° above the horizontal. After one more second it is moving horizontally. If g = 10ms^-2, the velocity of projection is