A projectile is shot from the ground at an angle of #pi/12 # and a speed of #3/5 m/s#. When the projectile is at its maximum height, what will its distance from the starting point be?

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Answer 1 · 2017-03-24T08:03:04

The distance is #=0.009m#

Solving in the vertical direction #uarr^+#

#u=u_0sintheta=3/5sin(1/12pi)#

#a=-g#

#v=0# at the maximum height

We apply the equation

#v=u+at#

#t=(v-u)/a=(0-3/5sin(1/12pi))/(-g)#

#=3/(5g)sin(1/12pi)=0.016s#

Time to reach the maximum height is #=0.016s#

Solving in the horizontal direction #->+#

distance #d=u_0tcos(1/12pi)=0.016*3/5*cos(1/12pi)#

#=0.009m#