Question

If a projectile is shot at a velocity of `13 m/s` and an angle of `pi/6`, how far will the projectile travel before landing?

Answer 1

Vel of projection v =`13ms^-1`
Angle of projection = `theta=pi/6`
Horizontal component of vel of projection =`vcostheta=13cos(pi/6)ms^-1`
Time of flight T = `(2vsintheta)/g=(2*13*sin(pi/6))/g`
So horizontal distance covered before landing
`=vcostheta*T`
`=13^2/g*2sin(pi/6)cos(pi/6)`
`169/10*sqrt3/2=14.64m`
[`g = 10ms^-2` considered]