A projectile is shot from the ground at an angle of #pi/8 # and a speed of #5 /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?

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Answer 1 · 2016-11-18T15:25:48

The distance is #=0.014m#

The maximum height is when the vertical component of the velocity #=0#

We use the equation,

#v=u- g*t #

#v=0#

#u=u_0sintheta=5/8*sin(pi/8)#

#0=5/8sin(pi/8)-g*t#

#t=(5sin(pi/8))/((8g))#

This is the time to reach the maximum height.

To, get the horizontal distance from the starting point, we multiply by the horizontal component of the velocity.

#x=u_0costheta*t#

#x=5/8*cos(pi/8)*5sin(pi/8)/(8g)#

#x=(25/(64g))*1/2*sin(pi/4)#

#x=(25sqrt2)/(256g)=0.014m#