A projectile is shot from the ground at an angle of #pi/4 # and a speed of #3 /8 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?

1 Answer
Apr 1, 2017

The distance is #0.007m#

Explanation:

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=3/8*sin(1/4pi)#

Acceleration is #a=-g#

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=3/8sin(1/4pi)-g*t#

#t=3/8*1/g*sin(1/4pi)#

#=0.027s#

Resolving in the horizontal direction #rarr^+#

We apply the equation of motion

#s=u_x*t#

#=3/8cos(1/4pi)*0.027#

#=0.007m#