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The velocity of a projectile when it is at the greatest height is $\sqrt{\frac{2}{5}}$ times its velocity when it is at hal of its greatest height. Determine its angle of projection.
Step 1
Step 1: The velocity of a projectile at its greatest height is given by $v = u \cos \theta$, where $u$ is the initial velocity and $\theta$ is the angle of projection. Show more…
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The velocity of a projectile when it is at the greatest height is (sqrt (2//5)) times its velocity when it is at half of its greatest height. Determine its angle of projection
The velocity of a projectile, when it is at the greatest height, is $\sqrt{\frac{2}{5}}$ times its velocity when it is at half of its greatest height. The angle of projection is (A) $30^{\circ}$ (B) $45^{\circ}$ (C) $\tan ^{-1} \frac{2}{3}$ (D) $60^{\circ}$
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