Question

A projectile is shot from the ground at an angle of `(5 pi)/12` and a speed of `1/2 m/s`. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Answer 1

The distance is `=0.014m`

Explanation:

Resolving in the vertical direction `uarr^+`

initial velocity is `u_y=vsintheta=1/2*sin(5/12pi)`

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=1/2sin(5/12pi)-g*t`

`t=1/(2g)*sin(5/12pi)`

`=0.049s`

The greatest height is

`h=(1/2sin(5/12pi))^2/(2g)=0.012m`

Resolving in the horizontal direction `rarr^+`

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

`s=u_x*t`

`=1/2cos(5/12pi)*0.049`

`=0.006m`

The distance from the starting point is

`d=sqrt(h^2+s^2)`

`=sqrt(0.012^2+0.006^2)`

`=0.014m`