Question

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

Answer 1

`3.67m`

Explanation:

Let the velocity of projection of the projectile 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 xxT=(u^2sin(2alpha))/g`

In the given problem

`u=6m/s ,g=9.8m/s^2 and alpha=pi/4`

So the distance travaresed by the projectile before landing is

`R=(6^2*sin(2*pi/4))/9.8m~~3.67m`