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?

1 Answer
Jul 30, 2016

#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#