A projectile is shot from the ground at an angle of #pi/6 # and a speed of #5 /9 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

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Answer 1 · 2017-04-20T06:03:07

The distance is #=0.014m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=5/9*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=5/9sin(1/6pi)-g*t#

#t=5/(9g)*sin(1/6pi)#

#=0.028s#

Resolving in the horizontal direction #rarr^+#

To find the distance, we apply the equation of motion

#s=u_x*t#

#=5/9cos(1/6pi)*0.028#

#=0.014m#