Question

A projectile is shot at an angle of `(5pi)/12` and a velocity of `1 m/s`. How far away will the projectile land?

Answer 1

Range of projectile motion is expressed as `(u^2 sin (2theta)/g)`

Where `u` is the velocity of projection and `theta` is the angle of projection.

Given, `v=1 m"/"s` and `theta= (5pi)/12`

So, putting in the values, we get the range is `0.05 m`.