Question

A projectile is shot at an angle of `(5pi)/12` and a velocity of `17 m/s`. How far away will the projectile land?

Answer 1

`14.45m`

Explanation:

Let us consider that the `"Velocity of projection"=u=17m/s and "Angle of projection"=alpha=5pi/12`

The vertical component of the velocity =`usinalpha`

The horizontal component of the velocity =`ucosalpha`

Let us also consider that the time of flight of the projectile is T

This T time will be required to return the prrojectile to the ground i.e.after this "T"time net vertical displacement is zero

So we can write

`0=usinalpha*T-1/2*g*T^2`
`T=(2usinalpha)/g`

Horizontal displacement = distance from point of projection to the point of landing `="Horizotal component of velocity"xx T=ucosalphaxx(2usinalpha)/g=(u^2sin2alpha)/g=(17^2sin(2*5pi/12))/10m=(28.9*sin(pi-pi/6))=28.9*sin(pi/6)=28.9/2=14.45m`