Question

A projectile is shot from the ground at an angle of `pi/12` and a speed of `9 m/s`. When the projectile is at its maximum height, what will its distance from the starting point be?

Answer 1

`x=2,06422018349` `meters`

Explanation:

`theta =cancel(pi)/12*180/cancel(pi)=15^o`
`" t is the elapsed time from the starting to maximum point"`
`t=v_i*sin 15/g`
`t=(9*0,258819045103)/(9,81)`
`t=0,237448665232` `s`
`x=v_(ix)*t`
`v_(ix)=v_i*cos 15`
`v_(ix)=9*0,965925826289`
`v_(ix)=8,6933324366` `m/s`
`x=8,6933324366*0,237448665232`
`x=2,06422018349` `meters`