Question

A projectile is shot from the ground at an angle of `pi/6` and a speed of `3 /8 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

`x=0,00119 m`
`y=0,000266 m`

Explanation:

`theta =pi/6 *180 /pi`
`theta =30^o`
`t=v_i*sin theta/g " elapsed time to max height"`
`t=(3*sin 15)/(8*9,81)`
`t=(0,261)/(78,48)`
`t=0,0033 s`
`x=v_i*t*cos 15`
`x=3/8*0,0033 .cos 15`
`x=0,00119 m`
`y=v_i*t*sin15-1/2*g*t^2`
`y=3/8*0,0033*sin 15-1/2 * 9,81 *0,0033^2`
`y=0,00032-0,0000534`
`y=0,000266 m`