Question

A projectile is shot from the ground at a velocity of `3 m/s` at an angle of `pi/3`. How long will it take for the projectile to land?

Answer 1

The time is `=1.06s`

Explanation:

The equation describing the trajectory of the projectile in the `x-y` plane is

`y=xtantheta-(gx^2)/(2u^2cos^2theta)`

The initial velocity is `u=3ms^-1`

The angle is `theta=(1/3pi)rad`

The acceleration due to gravity is `g=9.8ms^-2`

The distance `y=0`

Therefore,

`xtan(1/3pi)-(9.8*x^2)/(2*3^2cos^2(pi/3))=0`

`1.732x-1.089x^2=0`

`x(1.732-1.089x)=0`

`x=0`, this is the starting point

`x=1.732/1.089=1.59m`

The time is `t=x/v_x=1.59/(3cos(pi/3))=1.06s`