A projectile is shot at an angle of #(5pi)/12 # and a velocity of # 17 m/s#. How far away will the projectile land?

1 Answer
May 3, 2016

#14.45m#

Explanation:

Let us consider that the #"Velocity of projection"=u=17m/s and "Angle of projection"=alpha=5pi/12#

The vertical component of the velocity =#usinalpha#

The horizontal component of the velocity =#ucosalpha#

Let us also consider that the time of flight of the projectile is T

This T time will be required to return the prrojectile to the ground i.e.after this "T"time net vertical displacement is zero

So we can write

#0=usinalpha*T-1/2*g*T^2#
#T=(2usinalpha)/g#

Horizontal displacement = distance from point of projection to the point of landing #="Horizotal component of velocity"xx T=ucosalphaxx(2usinalpha)/g=(u^2sin2alpha)/g=(17^2sin(2*5pi/12))/10m=(28.9*sin(pi-pi/6))=28.9*sin(pi/6)=28.9/2=14.45m#