A bottle rocket can shoot its projectile vertically to a height of 22.5 m. (You can ignore air resistance.)

Question

I found the appropriate angle to reach max horizontal range was 45, what would the max horizontal range be

Answer 1

Hi Sheridan,

You can solve this from the following steps

1. At 45^O the horizontal v_h and vertical v_v (at start) components ate equal

2. v_v=Sqrt(2*g*h_max) where h_max - max. height

3. The total time t_air in air is twice the time t_max of reaching the h_max where t_max=Sqrt(2h_max/g)

4

The range L=v_h t_air=2v_h t_max

_{When you combine these points you'd get the solution. I hope that would be helpful for you to solve not only this one but other projectile problems.}

_Best

_{Dr.Ariel Boroda}

Answer 2

v₀ = √2gh; x = v₀cosθt; y = v₀sinθt - gt²/2 = 0; t = 0(start time) , t = 2v₀sinθ/g; x = v₀cosθ·2v₀sinθ/g = v₀²sin2θ/g; max x = v₀²/g = 2gh/g = 2h = 45 m when θ = 45°