A projectile is shot from the ground at an angle of #pi/12 # and a speed of #8 /15 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

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Answer 1 · 2017-04-08T12:48:19

The distance is #=0.007m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=8/15*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=8/15sin(1/12pi)-g*t#

#t=8/15*1/g*sin(1/12pi)#

#=0.014s#

Resolving in the horizontal direction #rarr^+#

We apply the equation of motion

#s=u_x*t#

#=8/15cos(1/12pi)*0.014#

#=0.007m#