Question

If a projectile is shot at a velocity of `9 m/s` and an angle of `pi/6`, how far will the projectile travel before landing?

Answer 1

7.1m

Explanation:

Let the velocity of projection of the object be u with angle of projection `alpha` with the horizontal direction.
The vertical component of the velocity of projection is `usinalpha` and the horizontal component is `ucosalpha`

Now if the time of flight be T then the object will return to the ground after T sec and during this T sec its total vertical dosplacement h will be zero. So applying the equation of motion under gravity we can write
`h=usinalpha xxT+1/2gT^2 =>0=uxxT-1/2xxgxxT^2` where `g="acceleration due to gravity"`
`:.T=(2usinalpha)/g`
The horizontal displacement during this T sec is `R=ucosalpha xx T=(u^2sin(2alpha))/g`

In our problem `u=9m/s;alpha=pi/6 "and" g=9.8m/s^2`

So the distance travelled by the projectile before landing is

`"Distance"=(9^2sin(2*pi/6))/9.8=7.1m`