Question

A projectile is shot from the ground at an angle of `pi/6` and a speed of `12 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 `d ~~ 6.5 m`

Explanation:

Write the equations for y(t) and x(t):

`y(t) = (-4.9 m/s^2)t^2 + (12m/s)sin(pi/6)t`

`x(t) = (12m/s)cos(pi/6)t`

Using the t coordinate for the axis of symmetry for y(t):

`h = -b/(2a) = -((12m/s)sin(pi/6))/(2(-4.9 m/s^2)) ~~ 0.6s`

The distance, d, is:

`d = sqrt(((-4.9 m/s^2)(0.6s)^2 + (12m/s)sin(pi/6)0.6s)^2 + ((12m/s)cos(pi/6)0.6s)^2)`

`d ~~ 6.5 m`