A projectile is shot from the ground at an angle of #pi/8 # and a speed of #6 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 · 2017-04-24T09:51:47

The distance is #=1.52m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=6*sin(1/8pi)#

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=6sin(1/8pi)-g*t#

#t=6/g*sin(1/8pi)#

#=0.23s#

Resolving in the horizontal direction #rarr^+#

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

#s=u_x*t#

#=7cos(1/8pi)*0.23#

#=152m#