A projectile is shot from the ground at an angle of #pi/6 # and a speed of #1 /2 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?

Question

1319 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-11T12:18:58

The distance is #=0.012m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=1/2*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=1/2sin(1/6pi)-g*t#

#t=1/(2g)*sin(1/6pi)#

#=0.026s#

The greatest height is

#h=(1/2sin(1/6pi))^2/(2g)=0.003m#

Resolving in the horizontal direction #rarr^+#

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

#s=u_x*t#

#=1/2cos(1/6pi)*0.026#

#=0.011m#

The distance from the starting point is

#d=sqrt(h^2+s^2)#

#=sqrt(0.003^2+0.011^2)#

#=0.012m#