If a projectile is shot at a velocity of #18 m/s# and an angle of #pi/12#, how far will the projectile travel before landing?

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Answer 1 · 2017-02-06T08:27:09

The distance is #=16.53m#

Solving in the vertical direction #uarr^+#

#u=18sin(pi/12)#

#v=0#

#a=-g#

Time of flight to reach greatest height, #t#

We apply,

#v=u+at#

#0=18sin(pi/12) - g t #

#t=18/gsin(pi/12)#

The time to land is #=2t=36/gsin(pi/12)#

Solving in the horizontal direction #rarr^+#

#u=18cos(pi/12)#

Distance #s=u*2t=18cos(pi/12)*36/gsin(pi/12)#

#=18^2/g*sin(pi/6)=16.53m#