# If a projectile is shot at a velocity of #2 m/s# and an angle of #pi/6#, how far will the projectile travel before landing?

##### 1 Answer

#### Explanation:

The vertical component of velocity at firing will be

#2sin(pi/6) = 2(1/2) = 1 m/s#

The horizontal component will be

#2cos(pi/6) = 2sqrt(3)/2 = sqrt(3) m/s#

Now we use the equation

We know that the only acceleration in projectile motion (neglecting air resistance and friction) is due to gravity. Also, we know that

#0 = v_(y0)t + 1/2a_yt^2#

#0 = t + 1/2(-9.8)t^2#

#0 = -4.9t^2 + t#

#0 = t(-4.9t + 1)#

#t= 0 or 0.204#

So the projectile will hit the ground after

#x = 0 + sqrt(3)(0.204) + 1/2a_xt^2#

There will be no acceleration in this case (there never is in the horizontal direction).

Therefore,

#x= sqrt(3)(0.204)#

#0.35 # m

Hopefully this helps!