Question

For a given launch velocity, at which launch angle does the projectile undergo the maximum horizontal displacement

Answer 1

C 45 is the answer :)

Answer 2

45°

Step-by-step explanation:

The horizontal displacement of any projectile (assuming the it is launched from the same height as it arrives) is given by:

`X = (2*V^2)/(g)*cos\alpha  *sin\alpha`

As you can see it depends on both the sine and cosine of the launch angle.

When you graph that multiplication (cos(α) * sin(α)), you can see that it has a maximum peak at
`\alpha =45\°` (shown in the picture)

You could also derive the expression and find its maximum but I'll leave that for another time.