Question

A projectile is shot from the ground at an angle of `pi/8` and a speed of `4 /9 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

We're not given the projectile's coefficient of drag, which we'd need to include to take wind resistance into consideration. So, we ignore this.

the vertical component of the projectile's velocity is:

`sin(pi/8) * 4/9 m/s`

The projectile's vertical velocity, expressed as a function of t, will then be:

`v_t = 4/9 - 9.81t`

And at max height, this is zero, so we can solve for the time when this occurs:

`4/9 - 9.81t = 0`
`4/9 = 9.81t`
`t = 4/(9 * 9.81) = 0.0453 sec`

Distance travelled is the horizontal velocity multiplied by this time value. Horizontal velocity is the initial velocity times the cosine of the launch angle.

`d = 0.0453 * cos(pi/8) * 4/9`

`= .00791 m` - not very far!