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

Answer:

h = 112.5 feets

Step-by-step explanation:

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

$h=\dfrac{v_o^2}{16}\sin\theta\cos\theta$

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

So,

$h=\dfrac{v_o^2}{32}\sin2\theta$

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,

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

So, 112.5 feets is the maximum possible distance.