Question

A projectile is shot from the ground at an angle of `( pi)/3` and a speed of `7 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

`2.86m`

Explanation:

Split the initial velocity into its vertical (`u_v`) and horizontal (`u_h`) components using trigonometry:

`u = 7`
`theta = pi/3`

`u_h = 7cos(pi/3)`
`u_v = 7sin(pi/3)`

We're looking at when the object reaches maximum height, which occurs when the vertical velocity becomes zero - it stops moving up, but hasn't quite started moving down yet.

Write out suvat for the vertical components,

`s = ?`
`u = 7sin(pi/3)`
`v = 0`
`a = -9.81`
`t = ?`

Now we can find distance, `s`, by the equation

`v^2 = u^2 + 2as`

`s = (v^2-u^2)/(2a)`

and put in values we know,

`s = (0^2 - (7sin(pi/3))^2)/(2 * -9.81) = 1.873m`

We can also find `t` by

`v = u + at`

`t = (v-u)/a = (-7sin(pi/3))/-9.81 = 0.62s`

For the horizontal components we can write out suvat again,
`s = ?`
`u = 7cos(pi/3)`
`v = 7cos(pi/3)`
`a = 0`
`t = 0.62`

and use the equation

`s = ut = vt`

because the horizontal velocity doesn't change (if we ignore air resistance),

`s = 7cos(pi/3) xx 0.62 = 2.17m`

Now we have the horizontal and the vertical distances. These are at right angles to each other, so the overall distance of the object at its maximum height can be found by Pythagoras,

`s = sqrt(1.87^2 + 2.17^2) = 2.86m`