A projectile is shot from the ground at an angle of #pi/6 # and a speed of #1 /2 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 · 2017-07-11T12:18:58

The distance is #=0.012m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=1/2*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=1/2sin(1/6pi)-g*t#

#t=1/(2g)*sin(1/6pi)#

#=0.026s#

The greatest height is

#h=(1/2sin(1/6pi))^2/(2g)=0.003m#

Resolving in the horizontal direction #rarr^+#

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

#s=u_x*t#

#=1/2cos(1/6pi)*0.026#

#=0.011m#

The distance from the starting point is

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

#=sqrt(0.003^2+0.011^2)#

#=0.012m#