A projectile is shot from the ground at an angle of #pi/12 # and a speed of #4 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 · 2017-04-01T14:51:07

The distance is #=0.41m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=4*sin(1/12pi)#

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=4sin(1/12pi)-g*t#

#t=4*1/g*sin(1/12pi)#

#=0.106s#

Resolving in the horizontal direction #rarr^+#

We apply the equation of motion

#s=u_x*t#

#=4cos(1/12pi)*0.106#

#=0.41m#