A projectile is shot from the ground at an angle of #( pi)/3 # and a speed of #1 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 · 2017-04-07T13:01:52

The distance is #=0.044m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=1*sin(1/3pi)#

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=1sin(1/3pi)-g*t#

#t=1*1/g*sin(1/3pi)#

#=0.088s#

Resolving in the horizontal direction #rarr^+#

We apply the equation of motion

#s=u_x*t#

#=1cos(1/3pi)*0.088#

#=0.044m#