A projectile is shot from the ground at an angle of #pi/12 # and a speed of #1 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 · 2017-04-18T06:31:09

The distance is #=0.026m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=1*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=1sin(1/12pi)-g*t#

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

#=0.0264s#

Resolving in the horizontal direction #rarr^+#

To find the distance, we apply the equation of motion

#s=u_x*t#

#=1cos(1/12pi)*0.0264#

#=0.026m#