Question

A projectile is shot from the ground at an angle of `pi/4` and a speed of `2/5 ms^-1`. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Answer 1

`8.1mm`

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 maximum point be `T`, at the point the velocity will momentary be zero.

`{ (s=,"not required",m),(u=,2/5 sin(pi/4)=1/5sqrt(2),ms^-1),(v=,0,ms^-1),(a=,-g,ms^-2),(t=,T,s) :}`

So we can calculate `T` using `v=u+at`

`:. 0=1/5sqrt(2)-gT`
`:. gT=1/5sqrt(2)`
`:. T=sqrt(2)/(5g)`

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 `s` using `s=ut`

`s=2/5cos(pi/4)T`
`:. s=1/5sqrt(2)T`
`:. s=1/5sqrt(2)sqrt(2)/(5g)`
`:. s=2/(25g)`

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

`s=2/245 = 0.008163...= 0.0081 m`, or `8.1mm`