Question

If a projectile is shot at a velocity of `15 m/s` and an angle of `pi/12`, how far will the projectile travel before landing?

Answer 1

The distance is `=11.48m`

Explanation:

Resolving in the vertical direction `uarr^+`

initial velocity is `u_y=vsintheta=15*sin(1/12pi)`

Acceleration is `a=-g`

At the maximum height, `v=0`

We apply the equation of motion

`v=u+at`

to calculate the time to reach the greatest height

`0=15sin(1/12pi)-g*t`

`t=15/(g)*sin(1/12pi)`

`=0.396s`

To find the horizontal distance, we apply the equation of motion

`s=2u_x*t`

`=2*15cos(1/12pi)*0.396`

`=11.48m`