A projectile is shot from the ground at an angle of #pi/6 # and a speed of #7 /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 · 2017-07-04T18:32:19

The distance is #=0.03m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=7/9*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=7/9sin(1/6pi)-g*t#

#t=7/(9g)*sin(1/6pi)#

#=0.04s#

The greatest height is

#h=(7/9sin(1/6pi))^2/(2g)=0.01m#

Resolving in the horizontal direction #rarr^+#

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

#s=u_x*t#

#=7/9cos(1/6pi)*0.04#

#=0.03m#

The distance from the starting point is

#d=sqrt(h^2+s^2)#

#=sqrt(0.01^2+0.03^2)#

#=0.03m#