Question

A projectile is shot at an angle of `pi/4` and a velocity of `6 ms^-1`. How far away will the projectile land?

Answer 1

`3.67 \ m` (3 sf)

Explanation:

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

`{: (v=u+at, " where ", s="displacement "(m)), (s=ut+1/2at^2, , u="initial speed "(ms^-1)), (s=1/2(u+v)t, , v="final speed "(ms^-1)), (v^2=u^2+2as, , a="acceleration "(ms^-2)), (s=vt-1/2at^2, , t="time "(s)) :}`

Horizontal Motion

The projectile will move under constant speed (NB we can still use "suvat" equation with `a=0`).

Suppose that the projectile will travel a distance `x \ m` in time `T \ s`, we must resolve the initial speed in the horizontal direction:

# { (s=,x,m), (u=,6 cos (pi/4)=3sqrt(2),ms^-1), (v=,"Not Required",ms^-1), (a=,0,ms^-2), (t=,T,s) :} #

So applying `s=ut+1/2at^2` we get

`x = 3sqrt(2)T`

Vertical Motion

The projectile travels under constant acceleration due to gravity. Its displacement will be `0` at the point where the projectile is at ground level, leading to two solution `t=0` (trivial) and `t=T`. Considering upwards as positive, and again resolving the initial speed (vertically this time):

# { (s=,0,m), (u=,6 sin (pi/4)=3sqrt(2),ms^-1), (v=,0,ms^-1), (a=,-g,ms^-2), (t=,T,s) :} #

Applying `s=ut+1/2at^2` we have:

`0 = 3sqrt(2)T+1/2(-g)T^2`
`:. 0 = 6sqrt(2)T-gT^2`
`:. T(6sqrt(2)-gT) = 0`

Leading to two possible solution:

`{ (T = 0,,"(trivial solution)"),(6sqrt(2)-gT=0,,) :}`

Hence we have:

`6sqrt(2)-gT = 0 => T = (6sqrt(2))/g`

And using the result we established from the horizontal motion, we have:

`x = 3sqrt(2) * (6sqrt(2))/g`
`\ \ = 36/g`

If we use `g=9.8 \ ms^(-2)` we get:

`x = 3.67 \ m` (3 sf)