A projectile is shot from the ground at an angle of #(5 pi)/12 # and a speed of #7 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 · 2017-03-11T07:53:14

Distance #=1.25m#

Solving in the vertical direction #uarr^+#

#u=u_0sintheta=7sin(5/12pi)#

#a=-g#

#v=0# at the maximum height

We apply the equation

#v=u+at#

#t=(v-u)/a=(0-7sin(5/12pi))/(-g)#

#=7/gsin(5/12pi)=0.69s#

Time to reach the maximum height is #=0.69s#

Solving in the horizontal direction #->+#

distance #d=u_0tcos(5/12pi)=0.69*7*cos(5/12pi)#

#=1.25m#