Question

A projectile is shot from the ground at a velocity of `7 m/s` at an angle of `pi/12`. How long will it take for the projectile to land?

Answer 1

The time is `=0.37s`

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=7ms^-1`

The angle is `theta=(1/12pi)rad`

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

The distance `y=0`

Therefore,

`xtan(1/12pi)-(9.8*x^2)/(2*7^2cos^2(1/12pi))=0`

`0.268x-0.107x^2=0`

`x(0.268-0.107x)=0`

`x=0`, this is the starting point

`x=0.268/0.107=2.5m`

The time is `t=x/v_x=2.5/(7cos(1/12pi))=0.37s`