Question

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?

Answer 1

`0.35` meters

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 `y =y_0 + v_(y0)t + 1/2a_yt^2`

We know that the only acceleration in projectile motion (neglecting air resistance and friction) is due to gravity. Also, we know that `y = y_0 = 0`, letting `0` be the ground.

`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 `0.204` seconds. Now we use the same kinematic equation although this time in `x`.

`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` is simply

`x= sqrt(3)(0.204)`

`0.35` m

`:.`The projectile will land `0.35` meters away from the launch point.

Hopefully this helps!