Question

If a projectile is shot at a velocity of `15 m/s` and an angle of `pi/6`, how far will the projectile travel before landing?

Answer 1

The horizontal distance is `=19.88m`

Explanation:

The trajectory of a projectile in the `x-y` plane is given by the equation

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

The initial velocity is `u=15ms^-1`

The angle is `theta=pi/6`

The acceleration due to gravity is `g=9.8ms^-2`

The vertical distance travelled is `y=0`

Therefore,

`xtan(pi/6)-(9.8x^2)/(2*15^2cos^2(pi/6))=0`

`0.577x-0.029x^2=0`

`x(0.577-0.029x)=0`

`x=0` which is the initial point

`x=0.577/0.029=19.88m`

graph{0.577x-0.029x^2 [-0.96, 21.54, -4.6, 6.65]}