A projectile is shot from the ground at an angle of #pi/6 # and a speed of #3/4 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 · 2017-03-16T18:20:45

The distance is #=0.025m#

Solving in the vertical direction #uarr^+#

#u=u_0sintheta=3/4sin(1/6pi)#

#a=-g#

#v=0# at the maximum height

We apply the equation

#v=u+at#

#t=(v-u)/a=(0-3/4sin(1/6pi))/(-g)#

#=3/(4g)sin(1/6pi)=0.038s#

Time to reach the maximum height is #=0.038s#

Solving in the horizontal direction #->+#

distance #d=u_0tcos(1/8pi)=0.038*3/4*cos(1/6pi)#

#=0.025m#