Projection Under uniform Acceleration

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Question

A gun fires a bullet at a speed of 140 ms^–1. If the bullet is to hit a target at the same level as the gun and at 1 km distance, the angle of projection may be

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Solution

$\large R = \frac{{{u^2}\sin 2\theta }}{g} \Rightarrow 1000 = \frac{{140 \times 140 \times \sin 2\theta }}{{9.8}}$
$\large sin 2\theta = \frac{{98}}{{196}} = \frac{1}{2}$
$\large 2\theta = {30^0} \Rightarrow \theta = {15^0}(or){75^0}$

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Projection Under uniform Acceleration

Remember concepts with our Masterclasses.

A gun fires a bullet at a speed of 140 ms–1. If the bullet is to hit a target at the same level as the gun and at 1 km distance, the angle of projection may be

Ready to Test Your Skills?

free study material

A gun fires a bullet at a speed of 140 ms^–1. If the bullet is to hit a target at the same level as the gun and at 1 km distance, the angle of projection may be