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?

Question

1143 views around the world

You can reuse this answer
Creative Commons License

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#