Question

If a projectile is shot at a velocity of `52 m/s` and an angle of `pi/4`, how far will the projectile travel before landing?

Answer 1

`276 m`

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)) :}`

Vertical Motion
Motion under constant acceleration due to gravity, applied vertically upwards

Let the total time that the projectile takes to its return to the ground after launch be `T`, We would expect two solutions as obviously T=0 is one solution. The total displacement will be zero.

`{ (s=,0,m),(u=,52sin(pi/4)=26sqrt(2),ms^-1),(v=,"not required",ms^-1),(a=,-g,ms^-2),(t=,T,s) :}`

So we can calculate `T` using `s=ut+1/2at^2`

`:. 0 = 26sqrt(2)T + 1/2(-g)T^2`
`:. 1/2gT^2-26sqrt(2)T=0`
`:. T((gT)/2-26sqrt(2)) =0`
`:. T =0, (52sqrt(2))/g`

Horizontal Motion
Under constant speed (NB we can still use "suvat" equation with a=0). Thje projectile will be in in the air for the same time, `t =T`

So we can calculate the horizontal displacement `s` using `s=ut`

`s = 52cos(pi/4)T`
`:. s = (26sqrt(2))((52sqrt(2))/g)`
`:. s = 2704/g`

So using `g=9.8 ms^-2` we have.

`s=2704/9.8 = 275.918...= 276 m`