Question

The speed at the maximum height of a projectile is half of its initial speed of projection u. Its range is equal to

• A. $\frac{u^2}{2g}$
• B. $\frac{u^2}{g}$
• C. $\frac{\sqrt{3}{u}^2}{2g}$
• D. $\frac{\sqrt{3}{u}^2}{g}$

Answer 1

The correct option is (B) $\frac{\sqrt{3}{u}^2}{2g}$

Step 1, Given data

Initial projection velocity = u

The speed at maximum height is half of the initial speed

Step 2, Finding the range

We know,

At maximum height

$v=u\cos \theta =\frac{u}{2}\Rightarrow \cos \theta =\frac{1}{2}$

So,

The angle of projection must be ${60}^{\circ }$

Now for finding the range we can use the formula

Range = $\frac{u^2\sin 2\theta }{g}$ $=\frac{u^2\sin 2\times {60}^{\circ }}{g}$ $=\frac{\sqrt{3}{u}^2}{2g}$

Hence the range is $\frac{\sqrt{3}{u}^2}{2g}$