Projection Under uniform Acceleration

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Question

A stone is projected from the ground with a velocity of 14 ms^–1 one second later it clears a wall 2 m high. The angle of projection is (g = 10 ms^–2)

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Solution

$\large y = (u\sin \theta )t - \frac{1}{2}g{t^2}$
$\large 2 = (14\sin \theta )1 - \frac{1}{2} \times 10 \times 1$
$\large \sin \theta = \frac{1}{2} = \sin {30^0} \Rightarrow \theta = {30^0}$

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Projection Under uniform Acceleration

Remember concepts with our Masterclasses.

A stone is projected from the ground with a velocity of 14 ms–1 one second later it clears a wall 2 m high. The angle of projection is (g = 10 ms–2)

Ready to Test Your Skills?

free study material

A stone is projected from the ground with a velocity of 14 ms^–1 one second later it clears a wall 2 m high. The angle of projection is (g = 10 ms^–2)