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!