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?

1 Answer
May 11, 2016

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#