Question

If a projectile is shot at a velocity of `45 m/s` and an angle of `pi/6`, how far will the projectile travel before landing?

Answer 1

Range of projectile motion is given by the formula `R=(u^2 sin 2 theta)/g` where,`u` is the velocity of projection and `theta` is the angle of projection.

Given, `v=45 ms^-1,theta=(pi)/6`

So, `R=(45^2 sin ((pi)/3))/9.8=178.95m`

This is the displacement of the projectile horizontally.

Vertical displacement is zero,as it returned to the level of projection.

Answer 2

The projectile will travel `=178.94m`

Explanation:

The equation of the trajectory of the projectile in the `(x,y)` plane is

`y=xtantheta-(gx^2)/(2u^2cos^2theta)`

The initial velocity is `u=45ms^-1`

The angle is `theta=pi/6`

The acceleration due to gravity is `=9.8ms^-1`

When the projectile will land when

`y=0`

Therefore,

`xtantheta-(gx^2)/(2u^2cos^2theta)=xtan(pi/6)-(9.8x^2)/(2*45^2*cos^2(pi/6))=0`

`x(0.577-0.0032x)=0`

`x=0.577/0.0032`

`=178.94m`

graph{0.577x-0.0032x^2 [-6.2, 204.7, -42.2, 63.3]}