Question

assume the initial velocity is 60 feet/second. what is the maximum horizontal distance possible and at what angle does this occur

Answer 1

h = 112.5 feets

Explanation:

The equation for horizontal distance "h" in feet of a projectile with initial velocity v₀ and initial angle theta is given by :

`h=(v_o^2)/(16)\sin\theta\cos\theta`

We know that,
`2\sin\theta\cos\theta=\sin2\theta`

So,

`h=(v_o^2)/(32)\sin2\theta`

Now we need to find the maximum horizontal distance possible and at what angle does this occur.

For maximum distance angle should be 45 degrees. Som,

`h=(60^2)/(32)\sin2(45)\\\\h=112.5\ \text{feet}`

So, 112.5 feets is the maximum possible distance.