Explanation:
At the original height:
Δx = D
Δy = H
v₀ₓ = v₀
v₀ᵧ = 0
aₓ = 0
aᵧ = g
First, find the time it takes to land. In the y direction:
Δy = v₀ᵧ t + ½ aᵧ t²
H = (0) t + ½ (g) t²
t = √(2H/g)
Next, find the distance traveled. In the x direction:
Δx = v₀ₓ t + ½ aₓ t²
D = v₀ √(2H/g) + ½ (0) (2H/g)
D = v₀ √(2H/g)
We now have an equation for D in terms of v₀ and H.
So the new distance at the new height (h) and speed (2v₀) is:
D = 2v₀ √(2h/g)
We know this is equal to the first distance, so:
v₀ √(2H/g) = 2v₀ √(2h/g)
√(2H/g) = 2 √(2h/g)
2H/g = 4 (2h/g)
H = 4h
h = H/4
