Question

A projectile is shot from the ground at a velocity of `12 m/s` at an angle of `pi/6`. How long will it take for the projectile to land?

Answer 1

1.2s

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 canh write
`h=usinalphaxxT+1/2gT^2`
`=>0=uxxT-1/2xxgxxT^2`
`"where" g = "acceleration" "due to" gravity”`
`:.T=(2usinalpha)/g`
In our problem `u=12m/s and alpha=pi/6`

Hence `T=(2*12sin(pi/6))/9.8`
`=(2xx12xx1/2)/9.8=1.2s`