Question

Using the image, explain what information you are given and what you could solve for. Derive an equation that solves for the maximum height (H) of the cannonball and the maximum distance the cannon ball travels (R).

Answer 1

Answer:

h = (v₀²sin²θ)/2g

R = (v₀²sin2θ)/g

Explanation:

general equation is s = s₀ + v₀t + ½at²

if the firing point is origin and UP and RIGHT are positive directions, and if we ignore air resistance.

In the vertical direction, and remembering that gravity opposes the initial vertical velocity, the equation becomes

y = 0 + (v₀sinθ)t - ½(g)t²

y = (v₀sinθ)t - ½(g)t²

at maximum height h, vertical velocity is zero. The initial vertical velocity is reduced to zero in a time of

t = (v₀sinθ)/g

entering this value into our equation for y

h = (v₀sinθ)(v₀sinθ)/g - ½(g)((v₀sinθ)/g)²

h = (v₀²sin²θ)/g - ½(g)(v₀²sin²θ)/g²

h = (v₀²sin²θ)/g - ½(v₀²sin²θ)/g

h = (v₀²sin²θ)/2g

In the horizontal the equation becomes

x = 0 + (v₀cosθ)t + ½(0)t²

x = (v₀cosθ)t

As it will take as long to fall as it took the projectile to rise to the a•pex.

R = (v₀cosθ)(2)((v₀sinθ)/g)

R = 2(v₀²cosθsinθ)/g

trig identity sin2θ = 2cosθsinθ

R = (v₀²sin2θ)/g

The algorithm thinks the word "a•pex" is a swear word so I have to write it like I have... What an a•pex