Question

A projectile is shot at an angle of `pi/8` and a velocity of `43 m/s`. How far away will the projectile land?

Answer 1

The range is `=133.5m`

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=43ms^-1`

The angle is `theta=(1/8pi)rad`

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

The distance `y=0`

Therefore,

`xtan(1/8pi)-(9.8*x^2)/(2*43^2cos^2(pi/8))=0`

`0.414x-0.0031x^2=0`

`x(0.414-0.0031x)=0`

`x=0`, this is the starting point

`x=0.414/0.0031=133.5m`

graph{0.414x-0.0031x^2 [-0.01, 14.04, -1.347, 5.673]}