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Question: The velocity at the maximum height of a projectile is half of that of the initial velocity of projec...

The velocity at the maximum height of a projectile is half of that of the initial velocity of projection. Its range on the horizontal plane is
A. 3u22g\dfrac{{\sqrt 3 {u^2}}}{{2g}}
B. u23g\dfrac{{{u^2}}}{{3g}}
C. 3u22g\dfrac{{3{u^2}}}{{2g}}
D. 3u2g\dfrac{{3{u^2}}}{g}

Explanation

Solution

The velocity at maximum height is only in the x axis and is given by ucos⁡θu\cos \theta . Equate it as given in the question. After getting the angle, put it in the equation of the range in a horizontal projectile and you will get the final answer

Complete step by step answer:
Given that the velocity at the maximum height of a projectile is half of that of the initial velocity of projection.
The velocity at maximum height is given by v=ucos⁡θv = u\cos \theta
Therefore
ucos⁡θ=u2 ⇒cos⁡θ=12  u\cos \theta = \dfrac{u}{2} \\\ \Rightarrow \cos \theta = \dfrac{1}{2} \\\
Since the angle in horizontal projectile is acute angle,
∴θ=60∘\therefore \theta = {60^ \circ }
Range in a horizontal projectile is given by R=2u2sin⁡θcos⁡θgR = \dfrac{{2{u^2}\sin \theta \cos \theta }}{g}
Now we put the value of θ\theta in the equation of range
R=2u2sin⁡60∘cos⁡60∘g ⇒R=2u23212g ∴R=3u22g  R = \dfrac{{2{u^2}\sin {{60}^ \circ }\cos {{60}^ \circ }}}{g} \\\ \Rightarrow R = \dfrac{{2{u^2}\dfrac{{\sqrt 3 }}{2}\dfrac{1}{2}}}{g} \\\ \therefore R = \dfrac{{\sqrt 3 {u^2}}}{{2g}} \\\

So, the correct answer is “Option A”.

Additional Information:
Horizontal Projectile corresponds to 2D motion of a particle. Here the necessary acceleration is produced by gravity. The time period of this projectile is the time taken for the projectile to reach back the ground after being projected into the air. The maximum height is the highest perpendicular distance from the surface of the Earth. The range is the maximum horizontal distance of the particle.

Note:
Students should also know the equations for the velocity at any point, range, time period, maximum height and other special results. Students should also know how to use these results in other types of projectiles. This knowledge can also be used in other types of projectiles like oblique projectiles at some height and projectiles in inclined planes.