Question

The velocity at maximum height of a projectile is root 3/2 times initial velocity of projection (u) . Its range on horizontal plane is

Answer 1

Given :

The velocity at maximum height of a projectile is √3/2 times initial velocity of projection.

To Find :

Range of the projectile.

Solution :

❖ First of all, we need to find angle of projection.

$\sf:\implies\:Initial\:Velocity\times\dfrac{\sqrt3}{2}=Velocity\:at\:max.\:height$

$\sf:\implies\:u\times\dfrac{\sqrt3}{2}=u\:cos\theta$

$\sf:\implies\:cos\theta=\dfrac{\sqrt3}{2}$

$\sf:\implies\:\theta=cos^{-1}\dfrac{\sqrt3}{2}$

$:\implies\:\underline{\boxed{\bf{\purple{\theta=30^{\circ}}}}}$

♦ Range of projectile :

$\sf:\implies\:R=\dfrac{u^2\:sin2\theta}{g}$

• u denotes initial velocity
• R denotes range
• g denotes acceleration
• θ denotes angle of projection

$\sf:\implies\:R=\dfrac{u^2\:sin2(30^{\circ})}{g}$

$\sf:\implies\:R=\dfrac{u^2\:sin60^{\circ}}{g}$

$:\implies\:\underline{\boxed{\bf{\orange{R=\dfrac{\sqrt3u^2}{2g}}}}}$

Answer 2

Given :

The velocity at maximum height of a projectile is √3/2 times initial velocity of projection.

To Find :

Range of the projectile.

Solution :

❖ First of all, we need to find angle of projection.

$\sf:\implies\:Initial\:Velocity\times\dfrac{\sqrt3}{2}=Velocity\:at\:max.\:height$

$\sf:\implies\:u\times\dfrac{\sqrt3}{2}=u\:cos\theta$

$\sf:\implies\:cos\theta=\dfrac{\sqrt3}{2}$

$\sf:\implies\:\theta=cos^{-1}\dfrac{\sqrt3}{2}$

$:\implies\:\underline{\boxed{\bf{\purple{\theta=30^{\circ}}}}}$

♦ Range of projectile :

$\sf:\implies\:R=\dfrac{u^2\:sin2\theta}{g}$

u denotes initial velocity

R denotes range

g denotes acceleration

θ denotes angle of projection

$\sf:\implies\:R=\dfrac{u^2\:sin2(30^{\circ})}{g}$

$\sf:\implies\:R=\dfrac{u^2\:sin60^{\circ}}{g}$

$:\implies\:\underline{\boxed{\bf{\orange{R=\dfrac{\sqrt3u^2}{2g}}}}}$