Question

The velocity at the maximum height of a projectile is half of its initial velocity of projection u. Its range on the horizontal plane is

Answer 1

At maximum height
velocity = uCosθ
u/2 = uCosθ
θ = 60°

R = u² Sin(2θ) / g
= u² Sin(2 × 60°) / g
= u²√3 / (2g)

Range on horizontal plane is u²√3 / (2g)

Answer 2

The horizontal plane’s range is  $\bold{R=\frac{u^{2}}{2 g} \times \sqrt{3}}$

SOLUTION:

The object possesses a projectile motion.  

The initial velocity of the object is “u”

At maximum height,  

The value of Velocity $=u \cos \theta$

According to the given condition

The velocity at the maximum height becomes half of the initial projectile velocity “u”

So, we have

$\frac{u}{2}=u \cos \theta$

The value of angle is thus,$\theta=60^{\circ}$

Using the formula to calculate the range “R” of the projectile .

$R=\frac{u^{2} \sin 2 \theta}{g}$

Putting the value of θ $\begin{aligned} R &=\frac{u^{2} \sin 2 \times 60^{\circ}}{g} \\ R &=\frac{u^{2}}{2 g} \times \sqrt{3} \end{aligned}$