Question

A projectile is shot at an angle of `pi/4` and a velocity of `2 m/s`. How far away will the projectile land?

Answer 1

I got `"Distance" ~=0.408" metres"`
To give a full explanation of the calculation method proved too long. So I have given a summery of the method I used.

Explanation:

`color(blue)("Target: Time of flight x horizontal component of projectile")`

Let total time of flight be `T_f`

Find time of flight using

Vertical component of projectile`= 2xxsin(pi/4) -> 2/sqrt(2)`
The vertical travel distance (upwards) at `T_f -> 2/sqrt(2)xxT_f`

This must match the downward distance due to gravity at`" " T_f`.

This is the mean velocity (downwards) multiplied by flight time

`" "(9.81xxT_f)/2 xxT_f`

Equate them to each other and you have a quadratic that can be solved using the standard form equation.

`" I got "T_f~=0.2883` seconds
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Having found `T_f` you then substitute into the distance equation of the horizontal component of the projectile.

`2xxcos(pi/4)xxT_f" " ->" "(2T_f)/sqrt(2)`

`" Distance" ~=0.408" metres"`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Comment")`

`color(brown)("The "underline("actual")" height at any time "T_i" where "0<=T_i<=T_f)` is

`(2/sqrt(2)xxT_i )-((9.81xxT_i^2)/2 )`