Question

The equation of a projectile is y=root3 x -gx2/2,the velocity of projection is

Answer 1

Explanation:

y=√3x-g/2x^2

y=tanA-1/2*gx^2/u^2cos^2A

ON COMPRING THE EQUATIONS

TANA=√3

A=60

and

g/2*u^2cos^2A=g/2

u^2cos^2A=1

u^2cos^2(60)=1

u^2(1/2)^2=1

u^2*1/4=1

u^2=4

u=2

The angle of projectile is 60 and initial velocity is 2m/s.

Answer 2

Given:

$y = x \sqrt{3} - \dfrac{g {x}^{2} }{2}$

To find:

• Velocity of projection ?

Calculation:

First of all, let's see the general equation of a projectile:

In x axis :

$x = u \cos( \theta) t$

In Y axis :

$y = u \sin( \theta) t - \dfrac{1}{2} g {t}^{2}$

Putting the value of 't' in eq.(2) :

$\boxed{y = x \tan( \theta) - \dfrac{g {x }^{2} }{2 {u}^{2} { \cos}^{2}( \theta) } }$

Now, comparing this to the given equations :

$\implies \: \tan( \theta) = \sqrt{3}$

$\implies \: \theta = {60}^{ \circ}$

Now, we can say :

$2 {u}^{2} { \cos}^{2} ( \theta) = 2$

$\implies 2 {u}^{2} { \cos}^{2} ( {60}^{ \circ} ) = 2$

$\implies 2 {u}^{2} \times \dfrac{1}{4} = 2$

$\implies u = 2 \: m {s}^{ - 1}$

So, initial velocity is 2 m/s.