Question

A projectile is shot at an angle of `pi/8` and a velocity of `1 m/s`. How far away will the projectile land?

Answer 1

`d = 1^2/g * sin 45 = 1/(sqrt 2g) = 1/(10sqrt2) = sqrt 2/20 m`

Explanation:

We shall use the range formula

`v = 1m/s , theta = pi/8 =>2theta = pi/4 = 45^@`

`color(magenta)("Interesting note " 45^@ "is the optimal angle of launch")`

`d = 1^2/g * sin 45 = 1/(sqrt 2g) = 1/(10sqrt2) = sqrt 2/20 m`