Question

A body is projected horizontally from certain height. After 2s its direction of motion makes 30° to the horizontal the its initial velocity is
1) 2g/ rout 3
2) 2× rout 3
3) rout 3 × g / 2
4) none​

Answer 1

Given :

A body is projected horizontal from certain height. After 2s its direction of motion makes 30° to the horizontal.

To Find :

Initial velocity of the body.

Solution :

❖ This question is based on the concept of height to ground projectile motion.

• In this motion, horizontal component of velocity remains constant but vertical component of velocity changes with time as acceleration due to gravity acts on the body.

Let after time t, projectile path makes an angle of θ with horizontal direction.

$\dag\:\underline{\boxed{\bf{\green{tan\theta=\dfrac{v_y}{v_x}=\dfrac{gt}{u}}}}}$

where u denotes initial velocity.

By substituting the given values,

$\sf:\implies\:tan\:30^{\circ}=\dfrac{2g}{u}$

$\sf:\implies\:\dfrac{1}{\sqrt3}=\dfrac{2g}{u}$

$:\implies\:\underline{\boxed{\bf{\purple{u=2g\sqrt3\:ms^{-1}}}}}$

Answer 2

Given :

A body is projected horizontal from certain height. After 2s its direction of motion makes 30° to the horizontal.

To Find :

Initial velocity of the body.

Solution :

❖ This question is based on the concept of height to ground projectile motion.

In this motion, horizontal component of velocity remains constant but vertical component of velocity changes with time as acceleration due to gravity acts on the body.

Let after time t, projectile path makes an angle of θ with horizontal direction.

$\dag\:\underline{\boxed{\bf{\green{tan\theta=\dfrac{v_y}{v_x}=\dfrac{gt}{u}}}}}$

where u denotes initial velocity.

By substituting the given values,

$\sf:\implies\:tan\:30^{\circ}=\dfrac{2g}{u}$

$\sf:\implies\:\dfrac{1}{\sqrt3}=\dfrac{2g}{u}$

$:\implies\:\underline{\boxed{\bf{\purple{u=2g\sqrt3\:ms^{-1}}}}}$