Question

A projectile is shot from the ground at an angle of `pi/4` and a speed of `2 m/s`. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Answer 1

The horizontal distance is `=2/g=0.2m`

Explanation:

At the maximum height, the vertical component of the velocity is `=0`

The vertical component of the velocity `=usintheta=2*sin(pi/4)`

`=2*sqrt2/2=sqrt2`

as `cos(pi/4)=sqrt2/2`

The horizontal component of the velocity `=usintheta=2*cos(pi/4)`

`=2*sqrt2/2=sqrt2`

as `sin(pi/4)=sqrt2/2`

We use the equation, `v=u+at`

So, `0=sqrt2-gt`

`t=sqrt2/g`

To find the height, we use `v^2=u^2+2as`

`0=2-2gh`

`h=2/(2g)=1/g`

The horizontal distance is `=u_0*t=sqrt2*sqrt2/g=2/g`

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