A projectile is shot from the ground at an angle of #pi/4 # and a speed of #3 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 · 2017-04-18T08:19:46

The distance is #=0.459m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=3*sin(1/4pi)#

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=3sin(1/4pi)-g*t#

#t=3/g*sin(1/4pi)#

#=0.216s#

Resolving in the horizontal direction #rarr^+#

To find the distance, we apply the equation of motion

#s=u_x*t#

#=3cos(1/4pi)*0.216#

#=0.459m#