If a projectile is shot at a velocity of #15 m/s# and an angle of #pi/12#, how far will the projectile travel before landing?

1 Answer
Jun 12, 2017

The distance is #=11.48m#

Explanation:

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=15*sin(1/12pi)#

Acceleration is #a=-g#

At the maximum height, #v=0#

We apply the equation of motion

#v=u+at#

to calculate the time to reach the greatest height

#0=15sin(1/12pi)-g*t#

#t=15/(g)*sin(1/12pi)#

#=0.396s#

To find the horizontal distance, we apply the equation of motion

#s=2u_x*t#

#=2*15cos(1/12pi)*0.396#

#=11.48m#