A projectile is shot from the ground at an angle of #pi/6 # and a speed of #12 m/s#. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

Question

1328 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-27T05:48:49

The distance is #=6.36m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=12*sin(1/6pi)#

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=12sin(1/6pi)-g*t#

#t=12/g*sin(1/6pi)#

#=0.61s#

Resolving in the horizontal direction #rarr^+#

To find the horizontal distance, we apply the equation of motion

#s=u_x*t#

#=12cos(1/6pi)*0.61#

#=6.36m#