Question

If a projectile is shot at a velocity of `4 m/s` and an angle of `pi/6`, how far will the projectile travel before landing?

Answer 1

`1.412\ m`

Explanation:

The horizontal distance/ range `R` traveled by a projectile shot with an initial velocity `u=4\ \text{m/s` at an angle `\theta=\pi/6` with the horizontal is given by following formula

`R=\frac{u^2\sin2\theta}{g}`

`=\frac{4^2\sin(\pi/3)}{9.81}`

`=1.412\ m`

Answer 2

The distance is `=1.41m`

Explanation:

The trajectory of a projectile is given by the equation

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

Here,

The initial velocity is `u=4ms^-1`

The angle is `theta=pi/6`

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

Therefore,

`y=xtan(pi/6)-(9.8/(2*4^2cos^2(pi/6)))x^2`

`y=0.577x-0.408x^2`

The distance travelled horizontally is when `y=0`

`0=0.577x-0.408x^2`

`x(0.577-0.408x)=0`

`x=0`, this is the initial conditions

`x=0.577/0.408=1.41m`

graph{0.577x-0.408x^2 [-0.18, 2.858, -0.269, 1.249]}