A projectile is shot from the ground at an angle of #pi/8 # and a speed of #12 m/s#. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

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Answer 1 · 2017-02-02T08:40:19

The horizontal distance is #=5.2m#

Solving in the vertical direction #uarr^+#

#u=12sin(pi/8)ms^-1#

#v=0m s^-1#

#a=-g ms^-2#

#v=u+at#

#0=12sin(pi/8)-g t #

#t=(12sin(pi/8))/g#

This is the time to reach the greatest height

Solving in the horizontal direction #rarr^+#

#u=12cos(pi/8)#

#s=u*t=(12sin(pi/8))/g*12cos(pi/8)#

#=144/(g)*sin(pi/8)cos(pi/8)#

#=5.2m#