Question

A projectile is shot at an angle of `pi/8` and a velocity of `7 m/s`. How far away will the projectile land?

Answer 1

The projectile wiil land at `=3.54m`

Explanation:

The trajectory of a projectile is given by the equation

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

Here,

The initial velocity is `u=7ms^-1`

The angle is `theta=pi/8`

The acceleration due to gravity is `g=9.8ms^-1`

Therefore,

`y=xtan(pi/8)-(9.8/(2*7^2cos^2(pi/8)))x^2`

`y=0.414x-0.117x^2`

The distance travelled horizontally is when `y=0`

`0=0.414x-0.117x^2`

`x(0.414-0.117x)=0`

`x=0`, this is the initial conditions

`x=0.414/0.117=3.54m`

graph{0.414x-0.117x^2 [-0.18, 2.858, -0.269, 1.249]}