If a projectile is shot at a velocity of #6 m/s# and an angle of #pi/6#, how far will the projectile travel before landing?

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Answer 1 · 2017-04-21T08:26:47

The distance is #=1.02m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=6*sin(1/6pi)#

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=6sin(1/6pi)-g*t#

#t=6/g*sin(1/6pi)#

#=0.306s#

Resolving in the horizontal direction #rarr^+#

To find the distance where the projectile will land, we apply the equation of motion

#s=u_x*2t#

#=6cos(1/6pi)*0.106*2#

#=1.02m#