Question

If a projectile is shot at a velocity of `1 m/s` and an angle of `pi/3`, how far will the projectile travel before landing?

Answer 1

The distance is `=0.088m`

Explanation:

Resolving in the vertical direction `uarr^+`

initial velocity is `u_y=vsintheta=1*sin(1/3pi)`

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=1sin(1/3pi)-g*t`

`t=1/g*sin(1/3pi)`

`=0.088s`

Resolving in the horizontal direction `rarr^+`

To find the distance where the projectile will land, we apply the equation of motion

`s=u_x*2t`

`=1cos(1/3pi)*0.088*2`

`=0.088m`