A projectile is shot from the ground at an angle of #pi/6 # 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 · 2017-04-27T05:48:49

The distance is #=6.36m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=12*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=12sin(1/6pi)-g*t#

#t=12/g*sin(1/6pi)#

#=0.61s#

Resolving in the horizontal direction #rarr^+#

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

#s=u_x*t#

#=12cos(1/6pi)*0.61#

#=6.36m#