Question

A projectile is shot from the ground at an angle of `pi/8` and a speed of `1 /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

`Velocity_Y = 1/12 * sin(pi/8) = 0.0319 ms^-1`
`Velocity_X = 1/12 * cos(pi/8) = 0.0770 ms^-1`
Considering vertical movement, since at maximum height final velocity is `0 ms^-1`.

`U_Y = 0.0319ms^-1` `a_Y = -9.81 ms^-2` `V_Y = 0ms^-1`
We use `V = U +aT`,
`0 = 0.0319 + (-9.81)T`
`T = 0.00325 s`
Since horizontal component remains unchanged,
Horizontal Distance = `0.0770 * 0.00325 = 2.50` x `10^-4m`